TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 1.
If Q(h, k) is foot of the perpendicular from P(x1, y1) on the straight line ax + by + c = 0 then show that \(\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-\left(a x_1+b y_1+c\right)}{a^2+b^2}\). [May ’14, ’07; Mar. ’03]
Solution:
Let A(x1, y1), P(h, k)
‘P’ lies in ax + by + c = 0 then
ah + bk + c = 0
ah + bk = -c
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q1.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q1.2

Question 2.
Find the foot of the perpendicular from (-1, 3) on the straight line 5x – y – 18 = 0. [May ’13 (old), ’07: Mar. ’03]
Solution:
Given the equation of the straight line is 5x – y – 18 – 0
Comparing with ax + by + c = 0, we get
a = 5, b = -1, c = -18
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q2
Let the given point A(x1, y1) = (-1, 3)
Let (h, k) is the foot perpendicular from the point A(-1, 3) on line 5x – y – 18 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q2.1
∴ Foot of the perpendicular P(h, k) = (4, 2)

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 3.
If Q(h, k) is the image of the point p(x1, y1) with respect to the straight line ax + by + c = 0 then, show that \(\frac{h-\mathbf{x}_1}{\mathbf{a}}=\frac{\mathbf{k}-\mathbf{y}_1}{\mathbf{b}}=\frac{-2\left(a \mathbf{x}_1+\mathbf{b y}_1+\mathbf{c}\right)}{a^2+b^2}\). [Mar. ’19 (AP); Mar. ’13, ’04, ’93; May ’06, ’01]
Solution:
Let A(x1, y1), B(h, k)
‘C’ is the midpoint of \(\overline{\mathrm{AB}}\) then
C = \(\left(\frac{\mathrm{x}_1+\mathrm{h}}{2}, \frac{\mathrm{y}_1+\mathrm{k}}{2}\right)\)
‘B’ is the image of A, then midpoint ‘C’ lies on ax + by + c = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q3.1

Question 4.
Find the image of (1, -2) with respect to the straight line 2x – 3y + 5 = 0. [Mar. ’13]
Solution:
Given the equation of the straight line is 2x – 3y + 5 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = -3, c = 5
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q4
Let the given point A(x1, y1) = (1, -2)
Now, B(h, k) be the image of A(1, -2) with respect to the straight line 2x – 3y + 5 = 0
\(\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q4.1
∴ Image of B(h, k) = (-3, 4)

Question 5.
Find the equation of the straight lines passing through the point (-3, 2) and making an angle of 45° with the straight line 3x – y + 4 = 0. [Mar. ’09]
Solution:
Given, equation of the straight line is 3x – y + 4 = 0 ……..(1)
slope of the line (1) is m = \(\frac{-3}{-1}\) = 3
Let the given point A = (-3, 2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q5
Let the slope of the required line is m2 = m
∴ The equation of the required line is passed through (-3, 2), and having slope m is y – y1 = m(x – x1)
y – 2 = m(x + 3) …….(2)
y – 2 = mx + 3m
mx – y + (3m + 2) = 0
Given that the angle between the lines is 45°,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q5.1
Squaring on both sides
⇒ 10(m2 + 1) = 2(9m2 + 6m + 1)
⇒ 10m2 + 10 = 18m2 + 12m + 2
⇒ 8m2 + 12m – 8 = 0
⇒ 2m2 + 3m – 2 = 0
⇒ 2m2 + 4m – m – 2 = 0
⇒ (m + 2) (2m – 1) = 0
⇒ m + 2 = 0 (or) 2m – 1 = 0
⇒ m = -2 (or) \(\frac{1}{2}\)
If m = -2,
(2) ⇒ y – 2 = m(x + 3)
y – 2 = -2x – 6
2x + y + 4 = 0
If m = \(\frac{1}{2}\),
(2) ⇒ y – 2 = \(\frac{1}{2}\)(x + 3)
2y – 4 = x + 3
x – 2y + 7 = 0
∴ The equation of the straight line is 2x + y + 4 = 0
∴ The equation of straight line is x – 2y + 7 = 0
∴ The required equations of the straight line are, 2x + y + 4 = 0, x – 2y + 7 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 6.
Find the equations of the straight lines passing through the point of intersection of the lines 3x + 2y + 4 = 0, 2x + 5y = 1 and whose distance from (2, -1) is 2. [May ’09; Mar. ’09]
Solution:
Given equations of the straight lines are
3x + 2y + 4 = 0 ……….(1)
2x + 5y – 1 = 0 ………(2)
Let the given point A = (2, -1)
∴ The equation of the straight line passing through the point of intersection of lines is (1) & (2) is L1 + λL2 = 0
(3x + 2y + 4) + λ(2x + 5y – 1) = 0 ……(3)
3x + 2y + 4 + 2λx + 5λy – λ = 0
(3 + 2λ)x + (2 + 5λ)y + (4 – λ) = 0
Given that the perpendicular distance from point A(2, -1) to the line (3) is 2.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q6
squaring on both sides
(-λ + 4)2 = 29λ2 + 32λ + 13
λ2 + 16 – 8λ = 29λ2 + 32λ + 13
28λ2 + 40λ – 3 = 0
28λ2 + 42λ – 2λ – 3 = 0
14λ(2λ + 3) – 1(2λ + 3) = 0
(2λ + 3) (14λ – 1) = 0
(2λ + 3) = 0, 14λ – 1 = 0
λ = \(\frac{-3}{2}\); λ = \(\frac{1}{14}\)
Case I: If λ = \(\frac{-3}{2}\), then
The equation of the straight line is from (3)
(3x + 2y + 4) + (\(\frac{-3}{2}\)) (2x + 5y – 1) = 0
\(\frac{6 x+4 y+8-6 x-15 y+3}{2}=0\)
-11y + 11 = 0
y – 1 = 0
If λ = \(\frac{1}{14}\), then
The equation of the straight line is from (3)
(3x + 2y + 4) + (\(\frac{1}{14}\)) (2x + 5y – 1) = 0
\(\frac{42 x+28 y+56+2 x+5 y-1}{14}=0\)
44x + 33y + 55 = 0
4x + 3y + 5 = 0
∴ The required equations of the straight lines are y – 1 = 0 & 4x + 3y + 5 = 0

Question 7.
Find the orthocentre of the triangle whose vertices are (-5, -7), (13, 2), and (-5, 6). [Mar. ’16 (AP) ’12, ’03; May ’98]
Solution:
Let A(-5, -7), B(13, 2), C(-5, 6) are the given points.
The equation of the altitude \(\overline{\mathbf{A D}}\).
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q7
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q7.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q7.2

Question 8.
Find the circumcentre of the triangle whose vertices are (-2, 3), (2, -1) and (4, 0). [Mar. ’17 (AP), ’13 (Old), ’11; May ’15 (TS), ’02, ’92]
Solution:
Let A(-2, 3), B(2, -1), C(4, 0) are the given points
Let S(α, β) be the circumcentre of the triangle ABC
then, SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α + 2)2 + (β – 3)2 = (α – 2)2 + (β + 1)2
⇒ α2 + 4 + 4α + β2 + 9 – 6β = α2 + 4 – 4α + β2 + 1 + 2β
⇒ 4α + 9 – 6β + 4α – 1 – 2β = 0
⇒ 8α – 8β + 8 = 0
⇒ α – β + 1 = 0 …….(1)
Also SB = SC
⇒ SB2 = SC2
⇒ (α – 2)2 + (β + 1)2 = (α – 4)2 + (β – 0)2
⇒ α2 + 4 – 4α + β2 + 2β + 1 = α2 + 16 – 8α + β2
⇒ 4 – 4α + 2β + 1 – 16 + 8α = 0
⇒ 4α + 2β – 11 = 0 ……(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q8

Question 9.
Find the orthocentre of the triangle whose sides are given by x + y + 10 = 0, x – y – 2 = 0, and 2x + y – 7 = 0. [May ’13]
Solution:
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9.3

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 10.
Find the circumcentre of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0. [May ’11, ’05; Mar. ’06]
Solution:
Given, the equations of the straight lines are
3x – y – 5 = 0 ……..(1)
x + 2y – 4 = 0 ……..(2)
5x + 3y + 1 = 0 ……..(3)
Vertex A: Solving (1) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q10
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q10.1
Let S(α, β) be the circumcentre then SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β + 2)2 = (α – 2)2 + (β – 1)2
⇒ α2 – 2α + 1 + β2 + 4 + 4β = α2 + 4 – 4α + β2 + 1 – 2β
⇒ -2α + 4β + 4α + 2β = 0
⇒ 2α + 6β = 0
⇒ α + 3β = 0 ……..(4)
Also SB = SC
⇒ SB2 = SC2
⇒ (α – 2)2 + (β – 1)2 = (α + 2)2 + (β – 3)2
⇒ α2 – 2α + 1 + β2 – 2β + 1 = α2 + 4α + 4 + β2 – 6β + 9
⇒ 4α + 9 – 6β + 4α – 1 + 2β = 0
⇒ 8α – 4β – 8 = 0
⇒ 2α – β + 2 = 0 ……..(5)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q10.2

Question 11.
Find the equations of the straight lines passing through the point (1, 2) and make an angle of 60° with the line √3x + y + 2 = 0. [May ’03; B.P.]
Solution:
Given, equation of the straight line is √3x + y + 2 = 0 ……….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q11
Let the given point A(x1, y1) = (1, 2)
Given that θ = 60°
Let the slope of the required straight line is = m
∴ The equation of the straight line is passed through A(1, 2) and having slope ‘m’ is y – y1 = m(x – x1)
y – 2 = m(x – 1) …….(2)
y – 2 = mx – m
mx – y – m + 2 = 0
Let ‘θ’ be the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q11.1
Squaring on both sides
⇒ m2 + 1 = (√3m – 1)2
⇒ m2 + 1 = 3m2 + 1 – 2√3m
⇒ 2m2 – 2√3m = 0
⇒ m2 – √3m = 0
⇒ m(m – √3) = 0
⇒ m = 0 (or) m – √3 = 0
⇒ m = 0 (or) m = √3
Case 1: If m = 0 then, the equation of a required straight line is,
from (2), y – 2 = 0(x – 1)
y – 2 = 0
y = 2
Case 2: If m = √3 then, the equation of a required straight line is
from (2), y – 2 = √3(x – 1)
y – 2 = √3x – √3
√3x – y + 2 – √3 = 0
∴ The equations of the required straight lines are y = 2 and √3x – y + 2 – √3 = 0

Question 12.
If p and q are the lengths of the perpendiculars from the origin to the straight lines x sec α + y cosec α = a and x cos α – y sin α = a cos 2α prove that 4p2 + q2 = a2. [Mar. ’13 (Old), ’08; May ’13]
Solution:
Given, the equations of the straight lines are
x sec α + y cosec α – a = 0 …….(1)
x cos α – y sin α = a cos 2α ……….(2)
Let the point p(x1, y1) = (0, 0)
Now, p = the perpendicular distance from the point p(0, 0) to the straight line (1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q12
q = the perpendicular distance from the point p(0, 0) to the straight line (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q12.1
L.H.S = 4p2 + q2
= 4(a sin α cos α)2 + (a cos 2α)2
= 4a2 sin2α cos2α + a2 cos2
= a2(2 sin α cos α)2 + a2 cos2
= a2 sin2 2α + a2 cos2
= a2(sin2 2α + cos2 2α)
= a2
= R.H.S
∴ 4p2 + q2 = a2

Some More Maths 1B Straight Lines Important Questions

Question 13.
Find the equation of the straight line which makes an angle 135° with the positive X-axis measured counterclockwise and passing through the point (-2, 3).
Solution:
Given that, the inclination of a line is θ = 135°
The slope of a straight line is, m = tan θ
= tan 135°
= tan (180° – 45°)
= -tan 45°
= -1
Let the given point A(x1, y1) = (-2, 3)
∴ The equation of the straight line passing through the point (-2, 3) and having slope ‘-1’ is,
y – y1 = m(x – x1)
y – 3 = -1(x + 2)
y – 3 = -x – 2
x + y – 1 = 0

Question 14.
Find the equation of the straight line passing through the point (2, 3) and making non-zero intercepts on the axes of coordinates whose sum is zero. [Mar. ’12; May ’09]
Solution:
The equation of a straight line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1 ………(1)
Given that, the sum of the intercepts = 0
a + b = 0
b = -a
From (1), \(\frac{x}{a}+\frac{y}{-a}=1\)
x – y = a ……(2)
Since equation (2) passes through the point (2, 3) then,
2 – 3= a
∴ a = -1
Substitute the value of ‘a’ in equation (2)
x – y = -1
x – y + 1 = 0
∴ The equation is x – y + 1 = 0.

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 15.
Find the equation of the straight line passing through points (4, -3) and perpendicular to the line passing through points (1, 1) and (2, 3).
Solution:
Let A(1, 1), B(2, 3), C(4, -3) are the given points
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q3
Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{2}\)
∴ The equation of the straight line passing through C(4, -3) and having slope \(\frac{-1}{2}\) is y – y1 = \(\frac{1}{m}\) (x – x1)
y + 3 = \(\frac{-1}{2}\) (x – 4)
2y + 6 = -x + 4
x + 2y+ 6 – 4 = 0
x + 2y + 2 = 0

Question 16.
Transform the equation 3x + 4y + 12 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form.
Solution:
(i) Slope-intercept form:
Given, the equation of the straight line is 3x + 4y + 12= 0
4y = -3x – 12
y = \(\frac{-3 x}{4}-\frac{12}{4}=\left(\frac{-3}{4}\right) x+(-3)\)
which is in the form of y = mx + c
Slope, m = \(\frac{-3}{4}\), y-intercept, c = -3
(ii) Intercept form:
Given, the equation of the straight line is 3x + 4y + 12 =0
3x + 4y = -12 × 1
\(\frac{3 x}{-12}+\frac{4 y}{-12}=1\)
\(\frac{x}{-4}+\frac{y}{-3}=1\)
which is in the form of \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept, a = -4, y-intercept, b = -3
(iii) Normal form:
Given the equation of the straight line is, 3x + 4y + 12 = 0
3x + 4y= -12
-3x – 4y = 12
On dividing both sides with \(\sqrt{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q4
which is in the form x cos α + y sin α = p
∴ cos α = \(\frac{-3}{5}\), sin α = \(\frac{-4}{5}\), p = \(\frac{12}{5}\)

Question 17.
Find the distance between the parallel straight lines 5x – 3y – 4 = 0, 10x – 6y – 9 = 0. [Mar. ’13(old), ’09, ’02; May ’12]
Solution:
Given, the equations of the straight lines are
5x – 3y – 4 = 0
10x – 6y – 9 = 0 …….(2)
2(5x – 3y – 4) = 0
10x – 6y – 8 = 0 ………(1)
Comparing (1) with ax + by + c1 = 0, we get
a = 10, b = -6, c1 = -8
Comparing (2) with ax + by + c2 = 0, we get
a = 10, b = -6, c2 = -9
Distance between the parallel lines =
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q5

Question 18.
Find the value of p, if the straight lines 3x + 7y – 1 = 0 and 7x – py + 3 = 0 are mutually perpendicular.
Solution:
Given, the equations of the straight lines are
3x + 7y – 1 = 0 ……..(1)
7x – py + 3 = 0 ……..(2)
Slope of the line (1) is m1 = \(\frac{-3}{7}\)
Slope of the line (2) is m2 = \(\frac{-7}{-p}=\frac{7}{p}\)
Since the given lines are perpendicular then,
m1 × m2 = -1
\(\frac{-3}{7} \times \frac{7}{p}\) = -1
p = 3

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 19.
Find the value of p, if the lines 3x + 4y = 5, 2x + 3y = 4, px + 4y = 6 are concurrent. [May ’06]
Solution:
Given, the equations of the straight lines are,
3x + 4y – 5 = 0 ……..(1)
2x + 3y – 4 = 0 ……..(2)
px + 4y – 6 = 0 …….(3)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q7
∴ The point of intersection of the straight lines (1) & (2) is (-1, 2)
Now, given lines are concurrent then, the point of intersection lies on the line
p(-1) + 4(2) – 6 = 0
-p + 8 – 6 = 0
-p + 2 = 0
p = 2

Question 20.
Find the value of p, if the lines 4x – 3y – 7 = 0, 2x + py + 2 = 0, 6x + 5y – 1 = 0 are concurrent.
Solution:
Given, the equations of the straight lines are,
4x – 3y – 7 = 0 …….(1)
2x + py + 2 = 0 ……..(2)
6x + 5y – 1 = 0 ……….(3)
Solving (1) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q8
∴ The point of intersection of the straight lines (1) & (3) is, (1, -1).
Since the given lines are concurrent, then the point of intersection (1, -1) lies on line (2).
2(1) + p(-1) + 2 = 0
2 – p + 2 = 0
4 – p = 0
p = 4

Question 21.
Show that the lines 2x + y – 3 = 0, 3x + 2y – 2 = 0 and 2x – 3y – 23 = 0 are concurrent and find the point of concurrent.
Solution:
Given, the equations of the straight lines are,
2x + y – 3 = 0 …….(1)
3x + 2y – 2 = 0 ……….(2)
2x – 3y – 23 = 0 ………(3)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q9
∴ The point of intersection of the straight lines (1) & (2) is (4, -5).
Now substitute the point (4, -5) in equation (3)
2(4) – 3(-5) – 23 = 0
8 + 15 – 23 = 0
23 – 23 = 0
0 = 0
∴ Given lines are concurrent.
∴ Point of concurrence = (4, -5).

Question 22.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line, when \(\overline{\mathbf{A B}}\) is divided in the ratio 2 : 3 at (-5, 2).
Solution:
The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).
Now, P(-5, 2) divides \(\overline{\mathbf{A B}}\) in the ratio 2 : 3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q10
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q10.1

Question 23.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line when \(\overline{\mathbf{A B}}\) is divided in the ratio 1 : 2 at (-5, 4).
Solution:
The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).
Now, P(-5, 4) divides \(\overline{\mathbf{A B}}\) in the ratio 1 : 2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q11
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q11.1

Question 24.
If non-zero numbers a, b, c are in harmonic progression, then show that the equation \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a family of concurrent lines and find the point of concurrency.
Solution:
Given that, a, b, c are in harmonic progression.
then, \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\frac{2}{b}-\frac{1}{a}=\frac{1}{c}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q12
bx + ay + 2a – b = 0
b(x – 1) + a(y + 2) = 0
(x – 1) + \(\frac{a}{b}\) (y + 2) = 0
This is of the form L1 + λL2 = 0
∴ \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a set of lines passing through the point of intersection of
L1 = x – 1 = 0 ……..(1)
L2 = y + 2 = 0 ………(2)
Solve (1) & (2)
from (1), x – 1 = 0 ⇒ x = 1
from (2) y + 2 = 0 ⇒ y = -2
∴ The point of concurrence = (1, -2).
∴ \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a set of concurrent lines.

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 25.
If a, b, c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.
Solution:
Given that, a, b, c are in arithmetic progression, then
2b = a + c
c = 2b – a
Now, ax + by + c = 0
ax + by + (2b – a) = 0
ax + by + 2b – a = 0
a(x – 1) + b(y + 2) = 0
x – 1 + \(\frac{b}{a}\) (y + 2) = 0
This is of the form L1 + λL2 = 0
∴ ax + by + c = 0 represents a set of lines passing through the point of intersection of
L1 = x – 1 = 0 …….(1)
L2 = y + 2 = 0 ……..(2)
Solve (1) & (2)
from (1), x – 1 = 0 ⇒ x = 1
from (2), y + 2 = 0 ⇒ y = -2
∴ The point of concurrence = (1, -2).
∴ ax + by + c = 0 represents a set of concurrent lines.

Question 26.
A straight line parallels the line y = √3x passes through Q(2, 3), and cuts the line 2x + 4y – 27 = 0 at p then, finds the length of PQ.
Solution:
Given the equation of the straight line is y = √3x
This is of the form y = mx
⇒ m = √3
⇒ tan θ = √3
⇒ θ = 60°
Since \(\overline{\mathrm{PQ}}\) is parallel to the straight line y = √3x
∴ Slope of \(\overline{\mathrm{PQ}}\) = √3
Inclination of a straight line \(\overline{\mathrm{PQ}}\), 0 = 60°
Given point Q(x1, y1) = (2, 3)
∴ The coordinates of P = (x1 + r cos θ, y1 + r sin θ)
= (2 + r cos 60°, 3 + r sin 60°)
= (2 + r(\(\frac{1}{2}\)), 3 + \(\frac{\sqrt{3} r}{2}\)) where |r| = PQ
Since the point, ‘p’ lies on the straight line 2x + 4y – 27 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q14

Question 27.
A straight line with slope 1 passes through Q(-3, 5) and meets the line x + y – 6 = 0 at P. Find the distance PQ.
Solution:
Given, equation of the straight line is x + y – 6 = 0 ……….(1)
The slope of the straight line m = 1
Given point Q(x1, y1) = (-3, 5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q15
∴ The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope ‘1’ and passes through the point Q(-3, 5) is (y – y1) = m(x – x1)
y – 5 = 1(x + 3)
y – 5 = x + 3
x – y + 8 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q15.1

Question 28.
A straight line through Q(2, 3) makes an angle \(\frac{3 \pi}{4}\) with the negative direction of the X-axis. If the straight line intersects the line x + y – 7 = 0 at P, find the distance PQ.
Solution:
Given, equation of the straight line is x + y – 7 = 0 ……..(1)
Given point Q(x1, y1) = (2, 3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q16
Given that the straight line makes an angle \(\frac{3 \pi}{4}\) with the ‘-ve’ direction of the X-axis, then the straight line makes an angle π – \(\frac{3 \pi}{4}\) = \(\frac{\pi}{4}\) = 45° with the positive direction of X-axis.
∴ The slope of the straight line is m = tan θ
= tan 45°
= 1
The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope ‘1’ and passing through the point Q(2, 3) is,
y – y1 = m(x – x1)
y – 3 = 1(x – 2)
y – 3 = x – 2
x – y + 1 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q16.1

Question 29.
Find the foot of the perpendicular drawn from (4, 1) upon the straight line 3x – 4y + 12 = 0. [May ’08; June ’02]
Solution:
Given the equation of the straight line is 3x – 4y + 12 = 0.
Comparing with ax + by + c = 0, we get
a = 3, b = -4, c = 12
Let the given point A(x1, y1) = (4, 1)
Let (h, k) is the foot perpendicular from the point A(4, 1) on the line 3x – 4y + 12 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q17
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q17.1

Question 30.
Find the foot of the perpendicular drawn from (3, 0) upon the straight line 5x + 12y – 41 = 0.
Solution:
Given the equation of the straight line is 5x + 12y – 41 = 0
Comparing with ax + by + c = 0, we get
a = 5, b = 12, c = -41
Let the given point A(x1, y1) = (3, 0)
Let (h, k) is the foot of the perpendicular from the point A(3, 0) on the line 5x + 12y – 41 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q18
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q18.1

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 31.
Find the image of the point (1, 2) in the straight line 3x + 4y – 1 = 0. [Mar. ’07, ’02, ’97; May ’04, ’97]
Solution:
Given the equation of the straight line is 3x + 4y – 1 = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 4, c = -1
Let the given point A(x1, y1) = (1, 2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q19
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q19.1

Question 32.
x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining points A, B. If A = (-1, -3) find the coordinates of B. [Mar. ’13 (Old)]
Solution:
Given, equation of the straight line is x – 3y – 5 = 0
Comparing with ax + by + c = 0, we get
a = 1, b = -3, c = -5
Let the given point A(x1, y1) = (-1, -3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q20
x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining the points A, B then, B(h, k) is the image of A(-1, -3) with respect to the straight line x – 3y – 5 = 0.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q20.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q20.2

Question 33.
If 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, -4) and (α, β), find α + β.
Solution:
Given, the equation of the straight line is 2x – 3y – 5 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = -3, c = -5
Let the given point A(x1, y1) = (3, -4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q21
2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining the points A, B then, B(α, β) is the image of A(3, -4) with respect to the straight line 2x – 3y – 5 = 0.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q21.1
∴ Image of A(3, 4) is (α, β) = (-1, 2)
∴ α + β = -1 + 2 = 1

Question 34.
Find the orthocentre of the triangle whose vertices are (-2, -1), (6, -1), and (2, 5). [May ’12; Mar. ’07, ’04]
Solution:
Let the given vertices are A(-2, -1), B(6, -1) & C(2, 5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q22
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q22.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q22.2

Question 35.
Find the orthocentre of the triangle whose vertices are (5, -2), (-1, 2), and (1, 4).
Solution:
Let the given points are A(5, -2), B(-1, 2) & C(1, 4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q23
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q23.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q23.2

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 36.
Find the circumcentre of the triangle whose vertices are (1, 3), (0, -2), and (-3, 1). [Mar. ’18 (AP); May ’06]
Solution:
Let A(1, 3), B(0, -2), C(-3, 1) are the given points.
Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q24
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β – 3)2 = (α – 0)2 + (β + 2)2
⇒ α2 – 2α + 1 + β2 + 9 – 6β = α2 + β2 + 4 + 4β
⇒ 4 + 4β + 2α – 9 + 6β – 1 = 0
⇒ 2α + 10β – 6 = 0
⇒ α + 5β – 3 = 0 ……(1)
Also SB = SC
⇒ SB2 = SC2
⇒ (α – 0)2 + (β + 2)2 = (α + 3)2 + (β – 1)2
⇒ α2 + β2 + 4 + 4β = α2 + 9 + 6α + β2 + 1 – 2β
⇒ 9 + 6α – 2β + 1 – 4 – 4β = 0
⇒ 6α – 6β + 6 = 0
⇒ α – β + 1 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q24.1

Question 37.
Find the circumcentre of the triangle whose vertices are (1, 3), (-3, 5), and (5, -1). [Mar. ’18 (TS)]
Solution:
Let A(1, 3), B(-3, 5), C(5, -1) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q25
Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β – 3)2 = (α + 3)2 + (β – 5)2
⇒ α2 – 2α + 1 + β2 + 9 – 6β = α2 + 9 + 6α + β2 + 25 – 10β
⇒ 6α + 25 – 10β – 1 + 2α + 6β = 0
⇒ 8α – 4β + 24 = 0
⇒ 2α – β + 6 = 0 ……(1)
Also SB = SC
⇒ SB2 = SC2
⇒ (α + 3)2 + (β – 5)2 = (α – 5)2 + (β + 1)2
⇒ α2 + 9 + 6α + 25 + β2 – 10β = α2 + 25 – 10α + β2 + 1 + 2β
⇒ 9 + 6α – 10β + 10α – 1 – 2β = 0
⇒ 16α – 12β + 8 = 0
⇒ 4α – 3β + 2 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q25.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q25.2
Circumcentre S = (-8, -10)

Question 38.
Find the circumcentre of the triangle whose vertices are (1, 0), (-1, 2), and (3, 2). [May ’13 (Old)]
Solution:
Let A(1, 0), B(-1, 2), C(3, 2) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q26
Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β – 0)2 = (α + 1)2 + (β – 2)2
⇒ α2 – 2α + 1 + β2 = α2 + 2α + 1 + β2 + 4 – 4β
⇒ 2α – 4β + 4 + 1 – 1 + 2α = 0
⇒ 4α – 4β + 4 = 0
⇒ α – β + 1 = 0 ……….(1)
Also SB = SC
⇒ SB2 = SC2
⇒ (α + 1)2 + (β – 2)2 = (α – 3)2 + (β – 2)2
⇒ α2 + 2α + 1 + β2 + 4 – 4β = α2 + 9 – 6α + β2 + 4 – 4β
⇒ 1 + 2α – 9 + 6α = 0
⇒ 8α – 8 = 0
⇒ α – 1 = 0
⇒ α = 1
Substitute α = 1 in (1), we get
1 – β + 1 = 0
⇒ -β = -2
⇒ β = 2
∴ Circumcentre S = (1, 2)

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 39.
If the equations of the sides of a triangle are 7x + y – 10 = 0, x – 2y + 5 = 0, and x + y + 2 = 0, find the orthocenter of the triangle. [May ’09, ’00, ’97]
Solution:
Given, the equations of the straight lines are
7x + y – 10 = 0 …….(1)
x – 2y + 5 = 0 …….(2)
x + y + 2 = 0 ……….(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27.3

Question 40.
Find the orthocentre of the triangle whose sides are given by 4x – 7y + 10 = 0, x + y = 5, and 7x + 4y = 15.
Solution:
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.5

Question 41.
Find the orthocentre of the triangle formed by the lines x + 2y = 0, 4x + 3y – 5 = 0 and 3x + y = 0. [Mar. ’10]
Solution:
Given, the equations of the straight lines are,
x + 2y = 0 ……..(1)
4x + 3y – 5z = 0 ……….(2)
3x + y = 0 ………(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.5

Question 42.
Find the circumcentre of the triangle formed by the straight lines x + y = 0, 2x + y + 5 = 0, and x – y = 2.
Solution:
Given, the equations of the straight lines are
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30.1
∴ Vertex C = (-1, -3)
Let S(α, β) be the circumcentre of the triangle then, SA = SB = SC
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30.2
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β + 1)2 = (α + 5)2 + (β – 5)2
⇒ α2 – 2α + 1 + β2 + 1 + 2β = α2 + 25 + 10α + β2 + 25 – 10β
⇒ 50 + 10α – 10β – 1 + 2α – 2β – 1 = 0
⇒ 12α – 12β + 48 = 0
⇒ α – β + 4 = 0 ……..(4)
Also SB = SC
⇒ SB2 = SC2
⇒ (α + 5)2 + (β – 5)2 = (α + 1)2 + (β + 3)2
⇒ α2 + 25 + 10α + β2 + 25 – 10β = α2 + 2α + 1 + β2 + 6β + 9
⇒ 25 + 25 + 10α – 10β – 1 = -2α – 9 – 6β = 0
⇒ 8α – 16β + 40 = 0
⇒ α – 2β + 5 = 0 ……..(5)
Solving (4) & (5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30.3
α = -3; β = 1
∴ Circumcentre S = (-3, 1)

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 43.
Find the circumcentre of the triangle formed by the straight lines x + y + 2 = 0, 5x – y – 2 = 0, and x – 2y + 5 = 0. [May ’08]
Solution:
Given, the equations of the straight lines are
x + y + 2 = 0 ………(1)
5x – y – 2 = 0 ………(2)
x – 2y + 5 = 0 ………(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31.1
Let S(α, β) be the circumcentre of the triangle then, SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α + 3)2 + (β – 1)2 = (α – 0)2 + (β + 2)2
⇒ α2 + 9 + 6α + β2 + 1 – 2β = α2 + β2 + 4 + 4β
⇒ 6α – 2β + 9 + 1 – 4 – 4β = 0
⇒ 6α – 6β + 6 = 0
⇒ α – β + 1 = 0 ……..(4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31.2
Also SB = SC
⇒ SB2 = SC2
⇒ (α + 0)2 + (β + 2)2 = (α – 1)2 + (β – 3)2
⇒ α2 + β2 + 4 + 4β = α2 – 2a + 1 + β2 – 6β + 9
⇒ 4 + 4β – 1 + 2α – 9 + 6β = 0
⇒ 2α + 10β – 6 = 0
⇒ α + 5β – 3 = 0 ……..(5)
Solving (4) & (5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31.3

Question 44.
Find the equations of the straight lines passing through the point (-10, 4) and make an angle θ with the line x – 2y = 10 such that tan θ = 2.
Solution:
Given, equation of the straight line is x – 2y – 10 = 0 ……..(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q32
Let the given point P(x1, y1) = (-10, 4)
Given that, tan θ = 2
cos θ = \(\frac{1}{\sqrt{5}}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q32.1
Let the slope of the required straight line = m.
∴ The equation of the straight line is passed through the point P(-10, 4) and having slope ‘m’ is y – y1 = m(x – x1)
y – 4 = m(x + 10)
y – 4 = mx + 10m
mx – y + 10m + 4 = 0
If ‘θ’ be the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q32.2
Squaring on both sides
5(m2 + 1) = 5(m + 2)2
5m2 + 5 = 5m2 + 20 + 20m (or) m = \(\frac{1}{0}\)
20m + 15 = 0
20m = -15
m = \(\frac{-3}{4}\) or m = \(\frac{1}{0}\)
Case 1: If m = \(\frac{-3}{4}\) then, the equation of the straight line is,
from (2), y – 4 = \(\frac{-3}{4}\) (x + 10)
4y – 16 = -3x – 30
3x + 4y – 16 + 30 = 0
3x + 4y + 14 = 0
Case 2: If m = \(\frac{1}{0}\) then, the equation of straight line is
from (2), y – 4 = \(\frac{1}{0}\) (x + 10)
0 = x + 10
x + 10 = 0
∴ Required equations of the straight lines are 3x + 4y + 10 = 0, x + 10 = 0

Question 45.
The base of an equilateral triangle is x + y – 2 = 0 and the opposite vertex is (2, -1). Find the equations of the remaining sides. [Mar. ’02]
Solution:
Given that base of an equilateral triangle is, x + y – 2 = 0 ……(1)
Let the opposite vertex is, A = (2, -1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q33
Since, Triangle ABC is an equilateral triangle B then,
A = B = C = 60°
Let the slope of the side \(\overline{\mathrm{AB}}\) = m
∴ The equation of the side \(\overline{\mathrm{AB}}\) passing through A(2, -1) and having slope ‘m’ is y – y1 = m(x – x1)
y + 1 = m(x – 2) ………(2)
y + 1 = mx – 2m
mx – y – 2m – 1 = 0
Let ‘θ’ be the angle between the lines (1) & (2) then,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q33.1
∴ The equations of the required straight lines are
from (2), y + 1 = 2 ± √3(x – 2).

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 46.
Prove that the points (1, 11), (2, 15), and (-3, -5) are collinear and find the equation of the straight line containing them.
Solution:
Let A(1, 11), B(2, 15), C(-3, -5) are the given points.
The equation of the straight line \(\overline{\mathrm{AB}}\) is,
(y – y1) (x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 11) (2 – 1) = (x – 1) (15 – 11)
⇒ (y – 11) (1) = (x – 1) (4)
⇒ y – 11 = 4x – 4
⇒ 4x – y + 7 = 0 ……….(1)
Now, substituting the point C(-3, -5) in equation (1)
4(-3) – (-5) + 7 = 0
⇒ -12 + 5 + 7 = 0
⇒ -7 + 7 = 0
⇒ 0 = 0
∴ The point C(-3, -5) lie on the straight line 4x – y + 7 = 0.
∴ Given points are collinear.
∴ The equation of the straight line is 4x – y + 7 = 0.

Question 47.
Find the angle which the straight line y = √3x – 4 makes with the Y-axis. [Mar. ’19 (AP & TS)]
Solution:
Given the equation of the straight line is y = √3x – 4
⇒ √3x – y – 4 = 0
Comparing the given equation with ax + by + c = 0, then
a = √3, b = -1, c = -4
The slope of a straight line √3x – y – 4 = 0 is
m = \(\frac{-a}{b}=\frac{-\sqrt{3}}{-1}\) = √3
tan θ = √3
∴ θ = 60° = \(\frac{\pi}{3}\)
The angle which the straight line y = √3x – 4 makes with X-axis = \(\frac{\pi}{3}\)
The angle which the straight line y = √3x – 4 makes with the Y-axis is = 180° – (90° + 60°)
= 180° – 150°
= 30°
= \(\frac{\pi}{6}\)

Question 48.
Write the equations of the straight lines parallel to the X-axis and (i) at a distance of 3 units above the X-axis and (ii) at a distance of 4 units below the X-axis.
Solution:
(i) The equation of the straight lines parallel to the X-axis, at a distance of 3 units above the X-axis is y = 3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q3
(ii) The equation of the straight line parallel to the X-axis at a distance of 4 units below the X-axis is y = -4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q3.1

Question 49.
Write the equations of the straight lines parallel to the Y-axis and (i) at a distance of 2 units from the Y-axis to the right of it (ii) at a distance of 5 units from the Y-axis to the left of it.
Solution:
(i) The equation of the straight line parallel to the Y-axis and at a distance of 2 units from the Y-axis to the right of it is x = 2.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q4
(ii) The equation of the straight line parallel to the Y-axis and at a distance of 5 units from the Y-axis to the left of it is x = -5.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q4.1

Question 50.
Find the equation of the straight line which makes an angle \(\frac{\pi}{4}\) with the positive X-axis in the positive direction and which passes through the point (0, 0).
Solution:
Given that, inclination of a straight line is θ = \(\frac{\pi}{4}\)
The slope of a line is m = tan θ
= tan \(\frac{\pi}{4}\)
= tan 45°
= 1
Let the given point A(x1, y1) = (0, 0)
∴ The equation of the straight line passing through A(0, 0) and having slope ‘1’ is y – y1 = m(x – x1)
⇒ y – o = 1(x – 0)
⇒ y = x – o
⇒ y = x
⇒ x – y = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 51.
Find the equation of the straight line which makes an angle of 135° with the positive X – axis in the positive direction and which pass through the point (3, -2).
Solution:
Given that, the inclination of a straight line is θ = 135°
The slope of a line is m = tan θ
= tan 135°
= tan (180° – 45°)
= tan 45°
= -1
Let the given point A(x1, y1) = (3, -2)
∴ The equation of the straight line passing through A(3, -2) and having slope ‘-1’ is y – y1 = m(x – x1)
⇒ y + 2 = -1(x – 3)
⇒ y + 2 = -x + 3
⇒ x + y + 2 – 3 = 0
⇒ x + y – 1 = 0

Question 52.
Find the equation of the straight line which makes an angle of 150° with the positive X-axis in the positive direction and the Y-intercept is 2.
Solution:
Given the inclination of a straight line is θ = 150°
The slope of a line is m = tan θ
= tan 150°
= -cot 60°
= \(\frac{-1}{\sqrt{3}}\)
y-intercept, c = 2.
The equation of the straight line having slope \(\frac{-1}{\sqrt{3}}\) and y-intercept ‘2’ is y = mx + c
⇒ y = \(\frac{-1}{\sqrt{3}}\)x + 2
⇒ y = \(\frac{-x+2 \sqrt{3}}{\sqrt{3}}\)
⇒ x + √3y – 2√3 = 0

Question 53.
Find the equation of the straight line which makes an angle \(\tan ^{-1}\left(\frac{2}{3}\right)\) with the positive X-axis in the positive direction and the y-intercept is 3.
Solution:
Given, inclination of a straight line is θ = \(\tan ^{-1}\left(\frac{2}{3}\right)\)
tan θ = \(\frac{2}{3}\)
Slope of a line is m = tan θ = \(\frac{2}{3}\)
y-intercept, c = 3
∴ The equation of a straight line having slope \(\frac{2}{3}\) and y-intercept ‘3’ is y = mx + c
⇒ y = \(\frac{2}{3}\) (x) + 3
⇒ y = \(\frac{2 x+9}{3}\)
⇒ 3y = 2x + 9
⇒ 2x – 3y + 9 = 0

Question 54.
Prove that the points (a, b + c), (b, c + a), (c, a + b) are collinear and find the equation of the straight line containing them.
Solution:
Let A(a, b + c), B(b , c + a), C(c, a + b) are the given points
The equation of the straight line \(\overline{\mathrm{AB}}\) is (y – y1) (x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – b – c) (b – a) = (x – a) (c + a – b – c)
⇒ (y – b – c) (b – a) = (x – a) (a – b)
⇒ (y – b – c) (b – a) = -(x – a) (b – a)
⇒ y – b – c = -x + a
⇒ x + y – a – b – c = 0 ……..(1)
Now, Substituting the point C(c, a + b) in equation (1)
x + y – a – b – c = 0
⇒ c + a + b – a – b – c = o
⇒ 0 = 0
∴ Point C(c, a + b) lies on the straight line x + y – a – b – c = 0.
∴ Given points are collinear.
∴ The straight line equation is x + y – a – b – c = 0.

Question 55.
A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equations of (i) \(\overline{\mathrm{AB}}\) (ii) the median through A (iii) the altitude through B (iv) the perpendicular bisector of the side \(\overline{\mathrm{AB}}\).
Solution:
(i) Equation of the straight line \(\overline{\mathrm{AB}}\) is (y – y1) (x2 – x1)= (x – x1)(y2 – y1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10
⇒ (y – 4)(-4 – 10) = (x – 10) (9 – 4)
⇒ (y – 4) (-14) = (x – 10) (5)
⇒ -14y + 56 = 5x – 50
⇒ 5x + 14y – 106 = 0
(ii) The equation of the median through ‘A’:
Since ‘D’ is the midpoint of BC, then
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10.1
The equation of the median through A is the equation of the straight line \(\overline{\mathrm{AD}}\),
(y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 4) (-3 – 10) = (x – 10)(4 – 4)
⇒ (y – 4) (-13) = (x – 10)(0)
⇒ (y – 4)(-13) = 0
⇒ y – 4 = 0
(iii) The equation of the altitude through ‘B’:
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10.2
= \(\frac{5}{12}\)
Since \(\overline{\mathrm{BE}} \perp \overline{\mathrm{AC}}\), then
slope of \(\overline{\mathrm{BE}}\) = \(\frac{-1}{m}=\frac{-1}{\frac{5}{12}}=\frac{-12}{5}\)
The equation of the altitude through B is, the equation of the straight line passing through B(-4, 9) and having slope \(\frac{-12}{5}\) is y – y1 = m(x – x1)
⇒ y – 9 = \(\frac{-12}{5}\) (x + 4)
⇒ 5y – 45 = -12x – 48
⇒ 12x + 5y + 3 = 0
(iv) The equation of the perpendicular bisector of side \(\overline{\mathrm{AB}}\):
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10.3
∴ The equation of the perpendicular bisector of \(\overline{\mathrm{AB}}\) is, the equation of the straight line passing through F(3, \(\frac{13}{2}\)) and having slope \(\frac{14}{5}\) is y – y1 = m(x – x1)
⇒ y – \(\frac{13}{2}\) = \(\frac{14}{5}\)(x – 3)
⇒ \(\frac{2 y-13}{2}=\frac{14}{5}(x-3)\)
⇒ 10y – 65 = 28x – 84
⇒ 28x – 10y – 19 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 56.
A straight line passing through A(1, -2) makes an angle \(\tan ^{-1}\left(\frac{4}{3}\right)\) with the positive direction of the X-axis in the anti-clockwise sense. Find the points on the straight line whose distance from A is 5.
Solution:
Given point (x1, y1) = A(1, -2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q11
The inclination of the straight line is θ = \(\tan ^{-1}\left(\frac{4}{3}\right)\)
tan θ = \(\frac{4}{3}\)
sin θ = \(\frac{4}{5}\)
cos θ = \(\frac{3}{5}\)
Distance, |r| = 5 units
Required point = (x1 + |r| cos θ, y1 + |r| sin θ)
= (x1 ± r cos θ, y1 ± r sin θ)
= (1 ± 5 . \(\frac{3}{5}\), -2 ± 5 . \(\frac{4}{5}\))
= (1 ± 3, -2 ± 4)
= (1 + 3, (-2) + 4), (1 – 3, -2 – 4)
= (4, 2), (-2, -6)

Question 57.
Find the sum of the squares of the intercepts of the line 4x – 3y = 12 on the axes of coordinates. [Mar. ’18 (AP)]
Solution:
Given, the equation of the straight line is 4x – 3y = 12
\(\frac{x}{3}+\frac{y}{-4}=1\) which is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept, a = 3, y-intercept, b = -4
Now, the Sum of the squares of the intercepts = a2 + b2
= 32 + (-4)2
= 9 + 16
= 25

Question 58.
The intercepts of a straight line on the axes of coordinates are a and b. If p is the length of the perpendicular drawn from the origin to this line, write the value of p in terms of a and b.
Solution:
The intercepts of a straight line on the coordinate axes are a, b.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q13
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q13.1
This is of the form x cos α + y sin α = p
∴ p = the length of the perpendicular drawn from the origin to the line = \(\frac{|a b|}{\sqrt{a^2+b^2}}\)

Question 59.
If the product of the intercepts made by the straight line x tan α + y sec α = 1 (0 ≤ α ≤ \(\frac{\pi}{2}\)) on the coordinate axes is equal to sin α, find α.
Solution:
Given, the equation of the straight line is x tan α + y sec α = 1
\(\frac{x}{\frac{1}{\tan \alpha}}+\frac{y}{\frac{1}{\sec \alpha}}=1\)
\(\frac{x}{\cot \alpha}+\frac{y}{\cos \alpha}=1\)
This is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = cot α
y-intercept (b) = cos α
Given that, a product of the intercepts is equal to sin α.
cot α × cos α = sin α
⇒ \(\frac{\cos \alpha}{\sin \alpha}\) × cos α = sin α
⇒ \(\frac{\cos ^2 \alpha}{\sin ^2 \alpha}\) = 1
⇒ cot2α = 1
⇒ tan2α = 1
⇒ tan α = 1
⇒ α = \(\frac{\pi}{2}\) (∵ 0 ≤ α ≤ \(\frac{\pi}{2}\))

Question 60.
A straight line passing through A(-2, 1) makes an angle of 30° with \(\overline{\mathrm{OX}}\) in the positive direction. Find the points on the straight line whose distance from A is 4 units.
Solution:
Given point A(x1, y1) = (-2, 1)
Inclination of straight line = 30°
distance |r| = 4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q15
∴ Required points = (x1 + |r| cos θ, y1 + |r| sin θ)
= (x1 ± r cos θ, y1 ± r sin θ)
= (-2 ± 4 cos 30°, 1 ± 4 sin 30°)
= (-2 ± 4\(\left(\frac{\sqrt{3}}{2}\right)\), 1 ± 4(\(\frac{1}{2}\)))
= (-2 ± 2√3 ,1 ± 2)
= (-2 + 2√3, 1 + 2), (-2 – 2√3, +1 – 2)
= (-2 + 2√3, 3), (-2 + 2√3, -1)

Question 61.
A straight line whose inclination with the positive direction of the X-axis measured in the anti-clockwise sense is \(\frac{\pi}{3}\) makes a positive intercept on the Y-axis. If the straight line is at a distance of 4 from the origin; find its equation.
Solution:
Given that, inclination of a straight line is θ = \(\frac{\pi}{3}\)
Let ‘l’ is the required straight line.
Now, ‘N’ is the foot of the perpendicular from the origin to the straight line ‘L’
∴ ∠XON = 150°
∴ a = 150°
Given that, p = 4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q16
The equation of the straight line ‘L’ in the normal form is x cos α + y sin α = p
⇒ x cos (150°) + y sin (150°) = 4
⇒ x cos(180 – 30) + y sin (180 – 30) = 4
⇒ -x cos 30° + y sin 30° = 4
⇒ \(-x\left(\frac{\sqrt{3}}{2}\right)+y\left(\frac{1}{2}\right)=4\)
⇒ -√3x + y = 8
⇒ √3x – y + 8 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 62.
Find the ratio in which the straight line 2x + 3y – 10 = 0 divides the join of the points (2, 3) and (2, 10).
Solution:
Given equation of the straight line is L = 2x + 3y – 10 = 0
Comparing the equation with ax + by + c = 0, we get
a = 2, b = 3, c = -10
Let the given points are A(x1, y1) = (2, 3) and B(x2, y2) = (2, 10)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q17

Question 63.
State whether (3, 2) and (-4, -3) are on the same side or opposite sides of the straight line 2x – 3y + 4 = 0.
Solution:
Given equation of the straight line is L = 2x – 3y + 4 = 0
Let the given points are A(x1, y1) = (3, 2) and B(x2, y2) = (-4, -3)
Now, l11 = L(3, 2)
= 2(3) – 3(2) + 4
= 6 – 6 + 4
= 4 > 0
l22 = L(-4, -3)
= 2(-4) – 3(-3) + 4
= -8 + 9 + 4
= 5 > 0
Since l11 0 and l22 > 0, the given points are on the same side of the straight line, L = 0.

Question 64.
Find the ratio’s in which (i) the X-axis and (ii) the Y-axis divide the line segment \(\overline{\mathrm{AB}}\) joining A(2, -3) and B(3, -6).
Solution:
Given points are A(x1, y1) = (2, -3) and B(x2, y2) = (3, -6)
(i) The X-axis divides the line segment AB in the ratio is,
-y1 : y2 = -(-3) : -6
= 3 : -6
= 1 : -2
= -1 : 2
(ii) Y-axis divides \(\overline{\mathrm{AB}}\) in the ratio is,
-x1 : x2 = -2 : 3

Question 65.
Find the equation of the straight line passing through the point of intersection of the lines x + y + 1 = 0 and 2x – y + 5 = 0 and containing the point (5, -2).
Solution:
Given, the equations of the straight lines are
x + y + 1 = 0 ……….(1)
2x – y + 5 = 0 ………..(2)
Let the given point be (5, -2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q20
∴ The point of intersection of lines (1) & (2) is A = (-2, 1)
Now, the equation of the straight line \(\overline{\mathrm{AB}}\) is (y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 1) (5 + 2) = (x + 2) (-2 – 1)
⇒ (y – 1)(7) = (x + 2)(-3)
⇒ 7y – 7 = -3x – 6
⇒ 3x + 7y – 1 = 0

Question 66.
Find the ratio in which the straight line 3x + 4y = 6 divides the join of the points (2, -1) and (1, 1).
Solution:
Given equation of the straight line is L = 3x + 4y – 6 = 0
Comparing the equation with ax + by + c = 0, we get
a = 3, b = 4, c = -6
Let the given points are A(x1, y1) = (2, -1) and B(x2, y2) = (1, 1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q21

Question 67.
Transform the equation (2 + 5k)x – 3(1 + 2k)y + (2 – k) = 0 into the form L1 + λL2 = 0 and find the point of concurrency of the family of straight lines represented by the equation.
Solution:
Given, equation is (2 + 5k)x – 3(1 + 2k)y + (2 – k) = 0
⇒ 2x + 5kx – 3y – 6ky + 2 – k = 0
⇒ (2x – 3y + 2) + k(5x – 6y – 1) = 0
This is of the form l1 + λl2 = 0
where l1 = 2x – 3y + 2 = 0 ……..(1)
l2 = 5x – 6y – 1 = 0 ……..(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q22
∴ Point of concurrence = (5, 4)

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 68.
Transform the equation (k + 1)x + (k + 2)y + 5 = 0, into the form L1 + λL2 = 0 and find the point of concurrency of the family of straight lines represented by the equation.
Solution:
Given, equation is (k + 1)x + (k + 2)y + 5 = 0
⇒ kx + x + ky + 2y + 5 = 0
⇒ (x + 2y + 5) + k(x + y) = 0
This is of the form l1 + λl2 = 0
where l1 = x + 2y + 5 = 0 ………(1)
l2 = x + y = 0 ……..(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q23
∴ Point of concurrence = (5, -5)

Question 69.
Find the area of the triangle formed by the straight line x – 4y + 2 = 0 and the coordinate axes.
Solution:
Given, equation of the straight line is x – 4y + 2 = 0 ………(1)
Comparing with ax + by + c = 0, we get
a = 1, b = -4, c = 2
∴ The area of the triangle formed by the straight line (1) & the co-ordinate axis = \(\frac{c^2}{2|a b|}=\frac{2^2}{2|1(-4)|}\) = \(\frac{1}{2}\)

Question 70.
Find the area of the triangle formed by the straight line 3x – 4y + 12 = 0 and the coordinate axes.
Solution:
Given, equation of the straight line is 3x – 4y + 12 = 0 ………(1)
Comparing with ax + by + c = 0, we get
a = 3, b = -4, c = 12
∴ The area of the triangle formed by the straight line and the co-ordinate axis = \(\frac{c^2}{2|a b|}=\frac{12^2}{2|3(-4)|}=\frac{12^2}{2|-12|}=\frac{144}{2(12)}\)
= 6 sq. units.

Question 71.
Find the equation of the straight line passing through the points (-1, 2) and (5, -1) and the area of the triangle formed by it with the axes of coordinates. [May ’15 (TS)]
Solution:
Let the given points are A(x1, y1) = (-1, 2) & B(x2, y2) = (5, -1)
The equation of the straight line passing through the points A(-1, 2) & B(5, -1) is (y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 2) (5 + 1) = (x + 1) (-1 – 2)
⇒ (y – 2)(6) = (x + 1)(-3)
⇒ 2y – 4 = -x – 1
⇒ x + 2y – 3 = 0 ……..(1)
Comparing equation (1) with ax + by + c = 0, we get
a = 1, b = 2, c = -3
∴ The area of the triangle formed by the straight line (1) and the co-ordinate axis = \(\frac{c^2}{2|a b|}=\frac{9}{2|1(2)|}=\frac{9}{2(2)}=\frac{9}{4}\) sq. units

Question 72.
Find the set of values of ‘a’ if the points (1, 2) and (3, 4) lie to the same side of the straight line 3x – 5y + a = 0.
Solution:
Given, equation of the straight line is 3x – 5y + a = 0 ……..(1)
Comparing (1) with ax + by + c = 0, we get
a = 3, b = -5, c = a
Let the given points are A(x1, y1) = (1, 2) & B(x2, y2) = (3, 4)
Given that, the points (1, 2) & (3, 4) lie on the same direction 3x – 5y + a = 0, then -l11 : l22 < 0
⇒ \(\frac{-\mathrm{L}_{11}}{\mathrm{~L}_{22}}\) < 0
⇒ \(\frac{-\left(a x_1+b y_1+c\right)}{\left(a x_2+b y_2+c\right)}\) < 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q27

Question 73.
Find the angle between the lines 2x + y + 4 = 0 and y – 3x = 7.
Solution:
Given, the equation of the straight lines is
2x + y + 4 = 0 ……..(1)
y – 3x – 7 = 0
⇒ 3x – y + 7 = 0 …….(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = 2, b1 = 1, c1 = 4
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = 3, b2 = -1, c2 = 7
If ‘θ’ is the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q28

Question 74.
Find the angle between the lines √3x + y + 1 = 0 and x + 1 = 0.
Solution:
Given, the equation of the straight lines are
√3x + y + 1 = 0 …….(1)
x + 1 = 0 ………(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = √3, b1 = 1, c1 = 1
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = 1, b2 = 0, c2 = 1
If ‘θ’ is the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q29

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 75.
Find the angle between the lines ax + by = a + b, a(x – y) + b(x + y) = 2b.
Solution:
Given, the equation of the straight lines are
ax + by = a + b
ax + by – a – b = 0 …….(1)
a(x – y) + b(x + y) = 2b
ax – ay + bx + by = 2b
(a + b)x + (-a + b)y – 2b = 0 ………(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = a, b1 = b, c1 = -a – b
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = a + b, b2 = b – a, c2 = -2b
If ‘θ’ is the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q30

Question 76.
Find the equation of a straight line perpendicular to the line 5x – 3y + 1 = 0 and pass through the point (4, -3). [Mar. ’15 (TS)]
Solution:
Given, the equation of the straight line is 5x – 3y + 1 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q31
Slope of the given line is m = \(\frac{-5}{-3}=\frac{5}{3}\)
Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{\frac{5}{3}}=\frac{-3}{5}\)
Let the given point A(x1, y1) = (4, -3)
Equation of the straight line passing through (4, -3) and having slope \(\frac{-3}{5}\) is,
y – y1 = \(\frac{-1}{m}\)(x – x1)
⇒ y + 3 = \(\frac{-3}{5}\)(x – 4)
⇒ 5y + 15 = -3x + 12
⇒ 3x + 5y + 3 = 0

Question 77.
Find the value of k, if the straight line 6x – 10y + 3 = 0 and kx – 5y + 8 = 0 are parallel.
Solution:
Given, the equation of the straight line is
6x – 10y + 3 = 0 …….(1)
kx – 5y + 8 = 0 ……(2)
Slope of the line (1) is m1 = \(\frac{-a}{b}=\frac{-6}{-10}=\frac{3}{5}\)
Slope of the line (2) is m2 = \(\frac{-k}{-5}=\frac{k}{5}\)
Since the given lines are parallel then m1 = m2
⇒ \(\frac{3}{5}\) = \(\frac{k}{5}\)
⇒ k = 3

Question 78.
(-4, 5) is a vertex of a square and one of its diagonals is 7x – y + 8 = 0. Find the equation of the other diagonal.
Solution:
Let ABCD be a square.
Let the given point be A(-4, 5).
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q33
Equation of the diagonal \(\overline{\mathrm{AC}}\) is, 7x – y + 8 = 0 is given
Slope of the diagonal, \(\overline{\mathrm{AC}}\) = 7
Since in a square diagonals \(\overline{\mathrm{AC}}\) & \(\overline{\mathrm{BD}}\) are perpendicular
then slope of the diagonal \(\overline{\mathrm{BD}}\) = \(\frac{-1}{m}=\frac{-1}{7}\)
∴ The equation of the diagonal \(\overline{\mathrm{BD}}\) is y – y1 = \(\frac{-1}{m}\)(x – x1)
⇒ y – 5 = \(\frac{-1}{7}\)(x + 4)
⇒ 7y – 35 = -x – 4
⇒ x + 7y – 31 = 0

Question 79.
A(-1, 1), B(5, 3) are opposite vertices of a square in the XY plane. Find the equation of the other diagonal (not passing through A, B) of the square.
Solution:
Let ABCD be a square.
Given that, opposite vertices of a square are A(-1, 1) and B(5, 3).
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q34
The equation of straight line passing through E & slope -3 is y – y1 = \(\frac{-1}{m}\)(x – x1)
⇒ y – 2 = -3(x – 2)
⇒ y – 2 = -3x + 6
⇒ 3x + y – 8 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 80.
Show that lines x – 7y – 22 = 0, 3x + 4y + 9 = 0 and 7x + y – 54 = 0 form a right angled isosceles triangle.
Solution:
Given, the equations of the straight lines are
x – 7y – 22 = 0 …..(1)
3x + 4y + 9 = 0 …….(2)
7x + y – 54 = 0 ………(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q35
Let ‘A’ be the angle between the lines (1) & (3) then,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q35.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q35.2
∴ A = 90°, B = 45°, C = 45°
∴ Given lines form a right angle isosceles triangle.

Question 81.
If ab > 0, find the area of the rhombus enclosed by the four straight lines ax ± by ± c = 0.
Solution:
Given, the equations of the straight lines are
ax + by + c = 0 ……..(1)
ax + by – c = 0 ………(2)
ax – by + c = 0 ……..(3)
ax – by – c = 0 ………(4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36.3

Question 82.
Find the area of the parallelogram whose sides are 3x + 4y + 5 = 0, 3x + 4y – 2 = 0, 2x + 3y + 1 = 0 and 2x + 3y – 7 = 0.
Solution:
Given, the equations of the straight lines are
3x + 4y + 5 = 0 …….(1)
3x + 4y – 2 = 0 ………(2)
2x + 3y + 1 = 0 ……….(3)
2x + 3y – 7 = 0 ………..(4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37.3

Question 83.
Find the incentre of the triangle whose vertices are (1, √3), (2, 0), and (0, 0).
Solution:
Given, A (x1, y1) = (1, √3), B(x2, y2) = (2, 0), C(x3, y3) = (0, 0)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q38
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q38.1

Question 84.
Find the incentre of the triangle whose sides are x = 1, y = 1, and x + y = 1.
Solution:
Given, the equation of the straight lines are
x = 1 …….(1)
y = 1 ………(2)
x + y = 1 ……..(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39
Vertex A:
Solving (1) & (3)
from (1), x = 1
from (3), x + y = 1
⇒ 1 + y = 1
⇒ y = 0
∴ Vertex A = (1, 0)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39.1
Vertex B:
Solving (1) & (2)
from (1), x = 1
from (2), y = 1
∴ Vertex B = (1, 1)
Vertex C:
Solving (2) & (3)
from (2), y = 1
from (3), x + y = 1
⇒ x + 1 = 1
⇒ x = 0
∴ Vertex C = (0, 1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39.3

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 85.
Find the values of k, if the angle between the straight lines kx + y + 9 = 0 and 3x – y + 4 = 0 is \(\frac{\pi}{4}\).
Solution:
Given equations are
kx + y + 9 = 0 ……..(1)
3x – y + 4 = 0 ………(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = k, b1 = 1, c1 = 9
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = 3, b2 = -1, c2 = 4
Given that, θ = \(\frac{\pi}{4}\)
Let ‘θ’ is the angle between the lines (1) & (2) then
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q40
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q40.1
Squaring on both sides
⇒ (k2 + 1)(10) = 2(3k – 1)2
⇒ 10k2 + 10 = 18k2 + 2 – 12k
⇒ 8k2 – 12k – 8 = 0
⇒ 2k2 – 3k – 2 = 0
⇒ 2k2 – 4k + k – 2 = 0
⇒ (k – 2)(2k + 1) = 0
⇒ k = 2 or k = \(\frac{-1}{2}\)

Question 86.
Find the equation of the straight line passing through the origin and also through the point of intersection of the lines 2x – y + 5 = 0 and x + y + 1 = 0.
Solution:
Given equations of the straight lines are
2x – y + 5 = 0 ………(1)
x + y + 1 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q41
The point of intersection of lines (1) & (2) is P(-2, 1).
The equation of the straight line passing through the point O(0, 0) & P(-2, 1) is (y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 0) (-2 – 0) = (x – 0) (1 – 0)
⇒ -2y = x
⇒ x + 2y = 0

Question 87.
Find the equation of the straight line perpendicular to the line 2x + 3y = 0 and pass through the point of intersection of the lines x + 3y – 1 = 0 and x – 2y + 4 = 0.
Solution:
Given equations of the straight lines are
x + 3y – 1 = 0 ……..(1)
x – 2y + 4 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q42
∴ The point of intersection of lines (1) & (2) is P(-2, 1).
Given equation of the straight line 2x + 3y = 0 ……..(3)
Slope of line is m = \(\frac{-2}{3}\)
Since the required line is perpendicular to the line (3),
Slope of required line is \(\frac{-1}{m}=\frac{-1}{\frac{-2}{3}}=\frac{3}{2}\)
The equation of the straight line passing through P(-2, 1) & slope \(\frac{3}{2}\) is y – y1 = \(\frac{-1}{m}\) (x – x1)
⇒ y – 1 = \(\frac{3}{2}\) (x + 2)
⇒ 2y – 2 = 3x + 6
⇒ 3x – 2y + 8 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 88.
Find the value of ‘a’ if the distances of the points (2, 3) and (-4, a) from the straight line 3x + 4y – 8 = 0 are equal.
Solution:
Given the equation of the straight line is 3x + 4y – 8 = 0
Let the given points be P = (2, 3), Q (-4, a)
Now, the perpendicular distance from the point P(2, 3) to the straight line 3x + 4y – 8 = 0 is
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q43
The perpendicular distance from the point Q(-4, a) to the straight line 3x + 4y – 8 = 0 is,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q43.1
Given that two perpendicular distances are equal then,
2 = \(\frac{|4 a-20|}{5}\)
⇒ |4a – 20| = 10
⇒ 2|2a – 10| = 10
⇒ 2a – 10 = ±5
⇒ 2a – 10 = 5; 2a – 10 = -5
⇒ 2a = 15; 2a = 5
⇒ a = \(\frac{15}{2}\); a = \(\frac{5}{2}\)
∴ a = \(\frac{15}{2}\) (or) \(\frac{5}{2}\)

Question 89.
Two adjacent sides of a parallelogram are given by 4x + 5y = 0 and 7x + 2y = 0 and one diagonal is 11x + 7y = 9. Find the equations of the remaining sides and the other diagonal.
Solution:
Given that, two adjacent sides of a parallelogram are given by
4x + 5y = 0 …….(1)
7x + 2y = 0 ………(2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.4
⇒ -6y – 8 = 21x – 35
⇒ 21x + 6y – 27 = 0
⇒ 7x + 2y – 9 = 0
The equation of the diagonal \(\overline{\mathrm{BD}}\) is,
(y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 0)(1 – 0) = (x – 0)(1 – 0)
⇒ y(1) = x(1)
⇒ y = x
⇒ x – y = 0
∴ Two adjacent sides of a parallelogram are 4x + 5y – 9 = 0, 7x + 2y – 9 = 0.
The equation of one of the diagonals is x – y = 0.

Question 90.
Find the incentre of the triangle formed by the straight lines x + 1 = 0, 3x – 4y = 5, and 5x + 12y = 27.
Solution:
Given, the equation of the straight lines are
x + 1 = 0 ………(1)
3x – 4y = 5 ………..(2)
5x + 12y = 27 ………..(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45
Vertex A:
Solving (1) & (3)
from (1), x + 1 = 0
⇒ x = -1
from (3), 5(-1) + 12y = 27
⇒ -5 + 12y = 27
⇒ 12y = 32
⇒ y = \(\frac{32}{12}\) = \(\frac{8}{3}\)
∴ Vertex A = (-1, \(\frac{8}{3}\))
Vertex B:
Solving (1) & (2)
from (1), x + 1 = 0
⇒ x = -1
from (2), 3(-1) – 4y = 5
⇒ -3 – 4y = 5
⇒ -4y = 8
⇒ y = -2
∴ Vertex B = (-1, -2)
Vertex C:
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.4

Question 91.
Find the equation of the straight line parallel to the line 3x + 4y = 7 and passing through the point of intersecting of the lines x – 2y – 3 = 0, x + 3y – 6 = 0. [Mar. ’16 (TS)]
Solution:
Given, the equation of the straight lines are
x – 2y – 3 = 0 …….(1)
x – 3y – 6 = 0 …….(2)
Solving (1) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q46
∴ The point of intersection of lines (1) & (2) is \(\mathrm{P}\left(\frac{21}{5}, \frac{3}{5}\right)\)
Given equation of the straight line is 3x + 4y = 7 …….(3)
Slope of line is m = \(\frac{-3}{4}\)
Since the required line is parallel to line 3.
Slope of required line = m = \(\frac{-3}{4}\)
∴ The equation of the straight line passing through \(\mathrm{P}\left(\frac{21}{5}, \frac{3}{5}\right)\) & Slope = \(\frac{-3}{4}\) is y – y1 = m(x – x1)
⇒ \(y-\frac{-3}{5}=\frac{-3}{4}\left(x-\frac{21}{5}\right)\)
⇒ \(\frac{5 y-3}{5}=\frac{-3}{4}\left(\frac{5 x-21}{5}\right)\)
⇒ 5y – 3 = \(\frac{-3}{4}\)(5x – 21)
⇒ 20y – 12 = -15x + 63
⇒ 15x + 20y – 75 = 0
⇒ 3x + 4y – 15 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 92.
Find the value of p, if the straight lines 3x + py – 1 = 0, 7x – 3y + 3 = 0, are mutually perpendicular. [Mar. ’19 (TS)]
Solution:
Given lines are
3x + py – 1 = 0 ……..(1)
7x – 3y + 3 = 0 ………(2)
Slope of (1) is m1 = \(\frac{-3}{p}\)
Slope of (2) is m2 = \(\frac{-7}{-3}\) = \(\frac{7}{3}\)
Since (1) & (2) are perpendicular then
m1 × m2 = -1
⇒ \(\frac{-3}{p} \times \frac{7}{3}=-1\)
⇒ \(\frac{-7}{p}\) = -1
⇒ p = 7

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 1.
Find the condition for the points (a, 0), (h, k), and (0, b), where ab ≠ 0, to be collinear. [Mar. ’10]
Solution:
Let A(a, 0), B(h, k), C (0, b) be the given points.
Slope of \(\overline{\mathrm{AB}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{k}-0}{\mathrm{~h}-\mathrm{a}}=\frac{\mathrm{k}}{\mathrm{h}-\mathrm{a}}\)
Slope of \(\overline{\mathrm{BC}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{b}-\mathrm{k}}{0-\mathrm{h}}=\frac{\mathrm{b}-\mathrm{k}}{-\mathrm{h}}\)
Since points A, B, C are collinear, then
Slope of \(\overline{\mathrm{AB}}\) = Slope of \(\overline{\mathrm{BC}}\)
\(\frac{k}{h-a}=\frac{b-k}{-h}\)
⇒ -hk = (h – a) (b – k)
⇒ -hk = bh – hk – ab + ak
⇒ bh + ak = ab
⇒ \(\frac{b h}{a b}+\frac{a k}{a b}=1\)
\(\frac{h}{a}+\frac{k}{b}=1\), which is the required condition.

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 2.
Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2. [Mar. ’18 (AP & TS); May ’12; B.P.]
Solution:
Let A = (2, 5), B = (x, 3) are the given points.
Given, Slope of \(\overline{\mathrm{AB}}\) = 2
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q2

Question 3.
Find the value of y, if the line joining the points (3, y) and (2, 7) is parallel to the line joining the points (-1, 4) and (0, 6). [Mar. ’17 (TS), ’14, ’08]
Solution:
Let A = (3, y), B = (2, 7), C = (-1, 4), and D = (0, 6) are the given points.
Given,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q3
Since, the lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are parallel then the
Slope of \(\overline{\mathrm{AB}}\) = Slope of \(\overline{\mathrm{CD}}\)
⇒ y – 7 = 2
⇒ y = 9

Question 4.
Find the equation of the straight line which make 150° with the X-axis in the positive direction and which pass through the point (-2, -1). [May ’04]
Solution:
Given that inclination of a straight line is θ = 150°
The slope of a line is, m = tan θ
= tan (150°)
= tan (90° + 60°)
= -cot 60°
= \(\frac{-1}{\sqrt{3}}\)
Let the given point A(x1, y1) is (-2, -1).
∴ The equation of the straight line passing through A(-2, -1) and having slope \(\frac{-1}{\sqrt{3}}\) is y – y1 = m(x – x1)
⇒ y + 1 = latex]\frac{-1}{\sqrt{3}}[/latex] (x + 2)
⇒ √3(y + 1) = -1(x + 2)
⇒ √3y + √3 = -x – 2
⇒ x + √3y + √3 = -2
⇒ x + √3y + √3 + 2 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 5.
Find the equations of the straight lines passing through the origin and making equal angles with the coordinate axes. [May ’05]
Solution:
Let l1, l2 are the equations of the straight lines passing through the origin and making equal angles with the co-ordinate axes
Case I: Inclination of a straight line l1 is θ = 45°
Slope of a line l1 is, m = tan θ = tan 45° = 1
Let the given point O = (0, 0)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q5
∴ The equation of a straight line l1 passing through O(0, 0) and having slope ‘1’ is y – y1 = m(x – x1)
y – 0 = 1(x – 0)
⇒ y = x
⇒ x – y = 0
Case II: Inclination of a straight line l2 is θ = 135°
The slope of a line l2 is, m = tan θ
= tan 135°
= tan (90° + 45°)
= -tan 45°
= -1
Let the given point O = (0, 0)
∴ The equation of a straight line l2 passing through O(0, 0) and having slope ‘-1’ is y – y1 = m(x – x1)
⇒ y – 0 = -1(x – 0)
⇒ y = -x
⇒ x + y = 0
∴ Required equations of the straight lines are x – y = 0; x + y = 0

Question 6.
Find the equation of the straight line passing through (-4, 5) and cutting off equal and non-zero intercepts on the coordinate axes. [Mar. ’15 (TS) ’13 (Old), ’07, ’00; May ’10, ’08]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
Given that, the straight line making equal intercepts on the co-ordinate axis, then a = b
From (1),
\(\frac{x}{a}+\frac{y}{a}=1\)
x + y = a ……….(2)
Since equation (2) passes through the point (-4, 5) then,
-4 + 5 = a
∴ a = 1
Substitute the value of ’a’ in equation (2)
∴ x + y = 1

Question 7.
Find the equation of the straight line passing through (-2, 4) and making non-zero intercepts whose sum is zero. [Mar. ’15 (AP), ’13; May ’15 (TS), ’02]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
Given that, the straight line-making intercepts whose sum is ‘0’
i.e., a + b = 0
b = -a
From (1)
\(\frac{x}{a}+\frac{y}{-a}=1\)
x – y = -a …….(2)
Since equation (2) passes through the point (-2, 4) then,
-2 – 4 = a
∴ a = -6
Substitute the value of ‘a’ in equation (2)
x – y = -6
x – y + 6 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 8.
Find the equation of the straight line passing through the point (3, -4) and making X and Y-intercepts which are in the ratio 2 : 3. [Mar. ’08]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
Given that, the ratio of intercepts = 2 : 3
X-intercept = 2a
Y-intercept = 3a
From (1),
\(\frac{x}{2 a}+\frac{y}{3 a}=1\)
\(\frac{3 x+2 y}{6 a}=1\)
3x + 2y = 6a ……..(2)
Since equation (2) passes through point (3, -4) then,
3(3) + 2(-4)= 6a
⇒ 9 – 8 = 6a
⇒ 6a = 1
⇒ a = \(\frac{1}{6}\)
Substitute the value of ‘a’ in equation (2)
3x + 2y = 6(\(\frac{1}{6}\))
∴ 3x + 2y = 1

Question 9.
Find the equation of the straight line passing through the points \(\left(a t_1^2, 2 at_1\right)\) and \(\left(a t_2^2, 2 at_2\right)\). [Mar. ’14. ’04; May ’15 (AP), ’00]
Solution:
Let A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) are the given points
The equation of the straight line passing through the points A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) is
(y – y1) (x2 – x1) = (x – x1) (y2 – y1)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q9

Question 10.
Find the equation of the straight line passing through A(-1, 3) and (i) parallel (ii) perpendicular to the straight line passing through B(2, -5) and C(4, 6). [May ’12; Mar. ’11]
Solution:
A(-1, 3), B(2, -5), C(4, 6) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q10
(i) Slope of the parallel line = m = \(\frac{11}{2}\)
∴ The equation of the straight line passing through A(-1, 3) and having slope \(\frac{11}{2}\) is y – y1 = m(x – x1)
y – 3 = \(\frac{11}{2}\)(x + 1)
⇒ 2y – 6 = 11x + 11
⇒ 11x – 2y + 17 = 0
(ii) Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{11 / 2}=\frac{-2}{11}\)
∴ The equation of the straight line passing through A(-1, 3) and having slope \(\frac{-2}{11}\) is y – y1 = \(\frac{-1}{m}\) (x – x1)
y – 3 = \(\frac{-2}{11}\) (x + 1)
⇒ 11y – 33 = -2x – 2
⇒ 2x + 11y – 31 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 11.
A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equation of the altitude through B. [May ’13 (Old)]
Solution:
Slope of \(\overline{\mathrm{AC}}\) is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q11
The equation of the altitude through B is, the equation of the straight line passing through B(-4, 9) and having slope \(\frac{-12}{5}\) is
y – y1 = m(x – x1)
y – 9 = \(\frac{-12}{5}\) (x + 4)
5y – 45 = -12x – 48
12x + 5y + 3 = 0

Question 12.
If the portion of a straight line intercepted between the axes of coordinates is bisected at (2p, 2q), write the equation of the straight line. [May ’90]
Solution:
Let a, b be the intercepts of a line.
∴ The line cuts the X-axis at A(a, 0), Y-axis at B(0, b)
Midpoint of \(\overline{\mathrm{AB}}\) is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q12
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q12.1

Question 13.
Find the angle made by the straight line y = -√3x + 3 with the positive direction of the X-axis measured in the counterclockwise direction. [May ’94]
Solution:
Given, the equation of the straight line is y = -√3x + 3
Comparing this equation with y = mx + c, We get
m = -√3 (∵ m = tan θ)
⇒ tan θ = -√3
⇒ tan θ = tan \(\frac{2 \pi}{3}\)
⇒ θ = \(\frac{2 \pi}{3}\)
∴ The angle made by the straight line is θ = \(\frac{2 \pi}{3}\)

Question 14.
Transform the equation √3x + y = 4 into (i) slope-intercept form (ii) intercept form (iii) normal form. [May ’16 (TS)]
Solution:
Given, the equation of the straight line is √3x + y = 4
(a) Slope-intercept form:
√3x + y = 4
y = -√3x + 4 which is of the form y = mx + c
where, slope (m) = -√3, y-intercept (c) = 4
(b) Intercept form:
Given, the equation of the straight line is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14
which is in the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = \(\frac{4}{\sqrt{3}}\), y-intercept (b) = 4
(c) Normal form:
Given the equation of the straight line is √3x + y = 4
On dividing both sides with
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14.1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14.2
which is in the form x cos α + y sin α = p
∴ α = \(\frac{\pi}{6}\), p = 2

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 15.
Transform the equation x + y + 1 = 0 into normal form. [Mar. ’17 (AP), ’08; May ’10; B.P.; Mar. ’18 (TS)]
Solution:
Given, equation of the straight line is x + y + 1 = 0
x + y = -1
– x – y = 1
On dividing both sides with
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q15

Question 16.
Transform the equation 4x – 3y + 12 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form.
Solution:
(a) Slope-intercept form:
Given equation of the line is 4x – 3y + 12 = 0
3y = 4x + 12
y = \(\frac{4 x+12}{3}=\left(\frac{4}{3}\right) x+4\)
which is in the form of y = mx + c
∴ Slope = \(\frac{4}{3}\), y-intercept = 4
(b) Intercept form:
Given equation is 4x – 3y + 12 = 0
\(\frac{x}{-3}+\frac{y}{4}=1\)
which is in the form of \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept = -3, y-intercept = 4
(c) Normal form:
Given the equation of the straight line is 4x – 3y = -12
-4x + 3y = 12
On dividing both sides by \(\sqrt{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q16

Question 17.
Transform the equation x + y – 2 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form. [Mar. ’12]
Solution:
(a) Slope-intercept form:
Given equation of the straight line is, x + y – 2 = 0
y = -x + 2
which is in the form of y = mx + c
∴ Slope, m = -1, y-intercept, c = 2
(b) Intercept form:
Given equation of the straight line is, x + y – 2 = 0
x + y = 2
\(\frac{x+y}{2}\) = 1
\(\frac{x}{2}+\frac{y}{2}\) = 1
which is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = 2, y-intercept (b) = 2
(c) Normal form:
Given equation of the straight line is, x + y – 2 = 0
x + y = 2
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q17
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q17.1

Question 18.
Transform the equation √3x + y + 10 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form. [May ’04]
Solution:
(a) Slope-intercept form:
Given the equation of the straight line is, √3x + y + 10 = 0
y = -√3x – 10 = -√3x + (-10)
which is in the form of y = mx + c
∴ Slope, m = -√3, y-intercept, c = -10
(b) Intercept form:
Given the equation of the straight line is √3x + y + 10 = 0
√3x + y = -10 × 1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18
(c) Normal form:
Given equation of the straight line is, √3x + y + 10 = 0
√3x + y = -10
-√3x – y = 10
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18.1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18.2

Question 19.
If the area of the triangle is formed by the straight lines, x = 0, y = 0, and 3x + 4y = a [a > 0] is ‘6’. Find the value of ‘a’. [May ’11, Mar. ’09, ’07]
Solution:
Given equations of the straight lines are (a > 0) 3x + 4y = a, x = 0 and y = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 4, c = -a
The area of the triangle formed by this line and the co-ordinate axis is equal to \(\frac{c^2}{2|a b|}\)
Given that area of the triangle = 6
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q19

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 20.
Find the value of p, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent. [Mar. ’17 (TS), ’13; May ’15 (TS)]
Solution:
Given, the equation of the straight lines
x + p = 0 ……..(1)
y + 2 = 0 ………(2)
3x + 2y + 5 = 0 ………(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q20
∴ Point of intersection of the lines (2) & (3) is (\(\frac{-1}{3}\), -2)
since given lines are concurrent, then, the point of intersection (\(\frac{-1}{3}\), -2) lies on (1)
x + p = 0
⇒ \(\frac{-1}{3}\) + p = 0
⇒ p = \(\frac{1}{3}\)

Question 21.
Find the ratio in which the straight line 2x + 3y = 5 divides the line joining the points (0, 0) and (-2, 1). [Mar. ’14]
Solution:
Given the equation of the straight line is L = 2x + 3y – 5 = 0
Comparing the equation with ax + by + c = 0, we get
a = 2, b = 3, c = 5
Let the given points are A(x1, y1) = (0, 0) and B(x2, y2) = (-2, 1)
Required ratio = \(\frac{-\left(a x_1+b {y}_1+c\right)}{a x_2+b y_2+c}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q21

Question 22.
Find the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0. [Mar. ’19 (AP); May ’13]
Solution:
Given, equations of the straight lines are 3x + 4y – 3 = 0, 6x + 8y – 1 = 0
6x + 8y – 6 = 0 …..(1)
6x + 8y – 1 = 0 …..(2)
Comparing (1) with ax + by + c1 = 0, we get
a = 6, b = 8, c1 = -6
Comparing (2) with ax + by + c2 = 0, we get
a = 6, b = 8, c2 = -1
Distance between the parallel lines =
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q22

Question 23.
Find the equation of ‘k’, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45°. [Mar. ’12, ’08, ’82; May ’11, ’02]
Solution:
Given, the equations of the straight lines are
4x – y + 7 = 0 ……..(1)
kx – 5y – 9 = 0 ……..(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = 4, b1 = -1, c1 = 7
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = k, b2 = -5, c2 = -9
Given that, θ = 45°
If ‘θ’ is the angle between the given lines then,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q23
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q23.1
Squaring on both sides
⇒ 17(k2 + 25) = 2(4k + 5)2
⇒ 17k2 + 425 = 2(16k2 + 25 + 40k)
⇒ 17k2 + 425 = 32k2 + 50 + 80k
⇒ 15k2 + 80k – 375 = 0
⇒ 3k2 + 16k – 375 = 0
⇒ 3k2 + 25k – 9k – 75 = 0
⇒ k(3k + 25) – 3(3k + 25) = 0
⇒ (3k + 25) (k – 3) = 0
⇒ 3k + 25 = 0; k – 3 = 0
⇒ 3k = -25; k = 3
⇒ k = 3 or \(\frac{-25}{3}\)

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 24.
Find the equation of the straight line parallel to the line 2x + 3y + 7 = 0 and pass through the point (5, 4). [Mar. ’13, ’03]
Solution:
Given, the equation of the straight line is 2x + 3y + 7 = 0
Given points (5, 4)
The equation of the straight line parallel to 2x + 3y + 7 = 0 is 2x + 3y + k = 0
Since equation (1) passes through the point (5, 4) then,
2(5) + 3(4) + k = 0
⇒ 10 + 12 + k = 0
⇒ 22 + k = 0
⇒ k = -22
∴ The required equation of the straight line is 2x + 3y – 22 = 0

Question 25.
Find the value of k, if the straight lines y – 3kx + 4 = 0 and (2k – 1)x – (8k – 1)y – 6 = 0 are perpendicular. [Mar. ’10]
Solution:
Given, the equations of the straight lines are
y – 3kx + 4 = 0 ………(1)
(2k – 1)x – (8k – 1)y – 6 = 0 ……….(2)
Slope of the line (1) is m1 = \(\frac{-(-3 k)}{1}\) = 3k
Slope of the line (2) is m2 = \(\frac{-(2 k-1)}{-(8 k-1)}=\frac{(2 k-1)}{(8 k-1)}\)
Since the given lines are perpendicular then m1 × m2 = -1
\(3 \mathrm{k}\left(\frac{2 \mathrm{k}-1}{8 \mathrm{k}-1}\right)\) = -1
⇒ 3k(2k – 1) = -1(8k – 1)
⇒ 6k2 – 3k = -8k + 1
⇒ 6k2 + 5k – 1 = 0
⇒ 6k2 + 6k – k – 1 = 0
⇒ 6k(k + 1) – 1(k + 1) = 0
⇒ (k + 1)(6k – 1) = 0
⇒ k + 1 = 0 (or) 6k – 1 = 0
⇒ k = -1 (or) k = \(\frac{1}{6}\)

Question 26.
Find the perpendicular distance from the point (3, 4) to the straight line 3x – 4y + 10 = 0. [Mar. ’16 (AP); May. ’15 (AP)]
Solution:
Given the equation of the straight line is 3x – 4y + 10 = 0
Comparing with ax + by + c = 0, we get
a = 3; b = -4, c = 40
Let the given point p(x1, y1) = (3, 4)
The perpendicular distance from the point (3, 4) to the line 3x – 4y + 10 = 0 is
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q26

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 27.
Find the slopes of the lines x + y = 0 and x – y = 0. [Mar. ’17 (AP)]
Solution:
Slope of x + y = 0 is \(\frac{-a}{b}=\frac{-1}{1}\) = -1
Slope of x – y = 0 is \(\frac{-a}{b}=\frac{-1}{-1}\) = 1

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 9 Hyperbolic Functions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions

I.

Question 1.
Show that \(t \tanh ^{-1}\left(\frac{1}{2}\right)=\frac{1}{2} \log _e^3\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 1

Question 2.
If cosh \(x=\frac{5}{2}\) find the values of
(i) cosh (2x)
(ii) sinh (2x)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 2

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

Question 3.
Prove that for any x ∈ R sinh (3x) =3 sinh (3x) + 3sinh x +4sin3h3 x.
Solution:
LHS : sinh(3x) = sinh (2x + x)
= sinh (2x) cosh x + cosh (2x) sinh x
= (2sinh x cosh x) cosh x + (1 + 2sinh2 x) slnh x
= 2 sinh x cosh2 x + sinh x + 2sinh3 x
= 2sinh x (1 + sinh2 x) + sinh x + 2sinh3 x
= 3sinh x + 4sinh3 x = R.H.S.

Question 4.
Prove that for any x∈R
\(\tanh 3 x=\frac{3 \tanh x+\tanh ^3 x}{1+3 \tanh ^2 x}\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 3

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

Question 5.
If cosh x = secθ then prove that \(\tanh ^2\left(\frac{x}{2}\right)=\tan ^2\left(\frac{\theta}{2}\right)\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 4
Question 6.
sinhx = 5 then show that \(x=\log _e(5+\sqrt{26})\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 5

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

II.

Question 1.
\(\theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\) and \(x=\log _{\mathrm{e}}\left(\cot \left(\frac{\pi}{4}+\theta\right)\right)\) that prove that
(i) cosh x = sec2θ
(ii) sin hx= – tan2θ.
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 6
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 7

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

Question 2.
Draw the graph of y = sinh x.
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 8

Question 3.
Draw the graph of y = cosh x.
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 9

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 1.
Transform the equation \(\frac{x}{a}+\frac{y}{b}=1\) into the normal form when a > 0 and b > 0. If the perpendicular distance of the straight line from the origin is p, deduce that \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\). [May ’08, ’04, ’02, ’97, ’95, ’90; Mar. ’07, ’02, ’00]
Solution:
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q1

Question 2.
Find the points on the line 3x – 4y – 1 = 0 which are at a distance of 5 units from the point (3, 2). [Mar. ’16 (AP); ’15 (AP); B.P.]
Solution:
Let the given point A(x1, y1) = (3, 2)
Given the equation of the straight line is 3x – 4y – 1 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q2
Distance |r| = 5
Required points = (x1 + |r| cos θ, y1 + |r| sin θ)
= (x1 ± r cos θ, y1 + r sin θ)
= (3 ± 5 . \(\frac{4}{5}\), 2 ± 5 . \(\frac{3}{5}\))
= (3 ± 4, 2 ± 3)
= (3 + 4, 2 + 3); (3 – 4, 2 – 3)
= (7, 5), (-1, 1)

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 3.
Find the value of k, if the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0 and 8x – 11y – 33 = 0 are concurrent. [May ’07, ’95; Mar. ’05, ’80; Mar. ’18 (TS)]
Solution:
Given, the equations of the straight lines are,
2x – 3y + k = 0 ……..(1)
3x – 4y – 13 = 0 ……..(2)
8x – 11y – 33 = 0 ………(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q3
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q3.1
∴ The point of intersection (11, 5) lies on (1)
2x – 3y + k = 0
⇒ 2(11) – 3(5) + k = 0
⇒ 22 – 15 + k = 0
⇒ 7 + k = 0
⇒ k = -7

Question 4.
If the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, then prove that a3 + b3 + c3 = 3abc. [Mar. ’19 (AP); Mar. ’08; May. ’00]
Solution:
Given, the equations of the straight lines are,
ax + by + c = 0 ……(1)
bx + cy + a = 0 ………(2)
cx + ay + b = 0 ……..(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q4
\(\frac{y}{b c-a^2}=\frac{1}{a c-b^2}\) ⇒ y = \(\frac{b c-a^2}{a c-b^2}\)
∴ Point of intersection of the straight lines (1) & (2) is \(\left(\frac{a b-c^2}{a c-b^2}, \frac{b c-a^2}{a c-b^2}\right)\)
Since the given lines are concurrent, then the point of intersection lies on line (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q4.1
⇒ c(ab – c2) + a(bc – a2) + b(ac – b2) = 0
⇒ abc – c3 + abc – a3 + abc – b3 = 0
⇒ 3abc – a3 – b3 – c3 = 0
⇒ a3 + b3 + c3 = 3abc

Question 5.
A variable straight line drawn through the point of intersection of the straight lines \(\frac{x}{a}+\frac{y}{b}=1\) and \(\frac{x}{a}+\frac{y}{b}=1\) meets the coordinate axes at A and B. Show that the locus of the midpoint of \(\overline{\mathbf{A B}}\) is 2(a + b) xy = ab(x + y). [May ’05]
Solution:
Given equations of the straight lines are,
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q5
The point of intersection of lines (1) & (2) is
C = \(\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)\)
The equation of the straight line \(\overline{\mathbf{A B}}\) is the intercept from \(\frac{x}{p}+\frac{y}{q}=1\) ……(3)
The straight line (3) meets the X-axis at A(p, 0), Y-axis at B(0, q).
Let Q(x1, y1) be any point on the locus.
Since Q(x1, y1) is the midpoint of \(\overline{\mathbf{A B}}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q5.1
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q5.2
then, \(y_1\left(\frac{a b}{a+b}\right)+\left(\frac{a b}{a+b}\right) x_1=2 x_1 y_1\)
\(\frac{a b y_1+a b x_1}{a+b}=2 x_1 y_1\)
ab(x1 + y1) = (a + b) 2x1y1
∴ The locus of the midpoint ‘Q’ of AB is 2(a + b) xy = ab(x + y)

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 6.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line when (p, q) bisects \(\overline{\mathbf{A B}}\). [May ’90]
Solution:
The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).
Now, C(p, q) bisects \(\overline{\mathbf{A B}}\), then C is the midpoint of \(\overline{\mathbf{A B}}\).
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q6
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q6.1

Question 7.
A triangle of area 24 sq. units is formed by a straight line and the coordinate axes in the first quadrant, find the equation of the straight line if it passes through (3, 4). [May ’07]
Solution:
Equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q7
Since equation (1) passes through the point (3, 4) then,
\(\frac{3}{a}+\frac{4}{b}=1\)
\(\frac{3 b+4 a}{a b}\) = 1
3b + 4a = ab
4a = ab – 3b
4a = b(a – 3)
b = \(\frac{4 a}{a-3}\) ……(2)
Given that, area of ΔOAB = 24 sq.units
\(\frac{1}{2}\)ab = 24
ab = 48
\(a\left(\frac{4 a}{a-3}\right)=48\)
a2 = 12(a – 3)
a = 12a – 36
a2 – 12a + 36 = 0
(a – 6)2 = 0
a = 6
from (2), b = \(\frac{4(6)}{6-3}\) = 8
The equation of the straight line is, from (1)
\(\frac{x}{6}+\frac{y}{8}\) = 1
\(\frac{4 x+3 y}{24}\) = 1
4x + 3y = 24

Question 8.
If 3a + 2b + 4c = 0, then show that the equation ax + by + c = 0, represents a family of concurrent straight lines and find the point of concurrency. [May ’10]
Solution:
Given that,
3a + 2b + 4c = 0
4c = -3a – 2b
c = \(\frac{-3 a-2 b}{4}\)
Now, ax + by + c = 0
ax + by + \(\left(\frac{-3 a-2 b}{4}\right)\) = 0
\(\frac{4 a x+4 b y-3 a-2 b}{4}\) = 0
4ax + 4by – 3a – 2b = 0
a(4x – 3) + b(4y – 2) = 0
(4x – 3) + \(\frac{b}{a}\) (4y – 2) = 0
This is of the form L1 + λL2 = 0
Here, ax + by + c = 0, represents a set of lines passing through the point of intersection of the lines
L1 = 4x – 3 = 0 …….(1)
L2 = 4y – 2 = 0 ………(2)
Solving (1) & (2)
From (1), 4x – 3 = 0
4x = 3
x = \(\frac{3}{4}\)
from (2), 4y – 2 = 0
4y = 2
y = \(\frac{1}{2}\)
∴ The point of concurrence = \(\left(\frac{3}{4}, \frac{1}{2}\right)\)
∴ ax + by + c = 0 represents a set of concurrent lines.
The point of concurrence = \(\left(\frac{3}{4}, \frac{1}{2}\right)\)

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 9.
Find the point on the straight line 3x + y + 4 = 0, which is equidistant from the points (-5, 6) and (3, 2). [Mar. ’13; Nov. ’98]
Solution:
Given, equation of the straight line is 3x + y + 4 = 0 …….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q9
Let the given points are A(-5, 6) & B(3, 2)
Let P(x, y) be a point on the straight line 3x + y + 4 = 0,
Given that, PA = PB
\(\sqrt{(x+5)^2+(y-6)^2}=\sqrt{(x-3)^2+(y-2)^2}\)
Squaring on both sides
(x + 5)2 + (y – 6)2 = (x – 3)2 + (y – 2)2
x2 + 25 + 10x + y2 + 36 – 12y = x2 + 9 – 6x + y2 – 4y + 4
10x – 12y + 61 = -6x – 4y + 13
16x – 8y + 48 = 0
2x – y + 6 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q9.1

Question 10.
A straight line through Q(√3, 2) makes an angle \(\frac{\pi}{6}\) with the positive direction of the X-axis. If the straight line intersects the line √3x – 4y + 8 = 0 at P, find the distance PQ. [Mar. ’19 (TS); Mar. ’04]
Solution:
Given equation of the straight line is √3x – 4y + 8 = 0 ……..(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q10
Given point Q(x1, y1) = (√3, 2)
Inclination of a straight line, θ = \(\frac{\pi}{6}\) = 30°
Slope of a straight line, m = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope \(\frac{1}{\sqrt{3}}\) and passing through the
point Q(√3, 2) is, y – y1 = m(x – x1)
y – 2 = \(\frac{1}{\sqrt{3}}\) (x – √3)
√3y – 2√3 = x – √3
x – √3y + √3 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q10.1

Question 11.
The line \(\frac{x}{a}-\frac{y}{b}=1\) meets the X-axis at P. Find the equation of the line perpendicular to the line at P. [May ’03]
Solution:
Given, the equation of the straight line is \(\frac{x}{a}-\frac{y}{b}=1\) …..(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q11
Since line (1) meets the X-axis at P.
Then y-coordinate = 0
\(\frac{x}{a}-\frac{0}{b}=1\)
x = a
∴ The coordinates of P = (a, 0)
The slope of the line (1) is m = \(\frac{\frac{-1}{a}}{\frac{-1}{b}}=\frac{b}{a}\)
Slope of the perpendicular line = \(\frac{-1}{b/a}=\frac{-a}{b}\)
∴ The equation of the line passing through P(a, 0) and having slope \(\frac{-a}{b}\) is,
y – y1 = \(\frac{-1}{m}\) (x – x1)
y – 0 = \(\frac{-a}{b}\) (x – a)
y = \(\frac{-a}{b}\) (x – a)
by = -ax + a2
ax + by – a2 = 0
which is the required equation of a straight line.

Question 12.
Find the equation of the line perpendicular to the line 3x + 4y + 6 = 0 and make an intercept -4 on the X-axis. [Mar. ’10]
Solution:
Given, equation of the straight line is 3x + 4y + 6 = 0 …….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q12
Slope of the line (1) is m = \(\frac{-3}{4}\)
Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{\left(\frac{-3}{4}\right)}=\frac{4}{3}\)
Given that, the required a straight line making an intercept -4 on X-axis. Then P = (-4, 0).
Equation of the straight line passing through P(-4, 0) and having slope \(\frac{4}{3}\) is
(y – y1) = \(\frac{-1}{m}\) (x – x1)
y – 0 = \(\frac{4}{3}\) (x + 4)
3y = 4x + 16
4x – 3y + 16 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 13.
Find the equation of the straight line making non-zero equal intercepts on the coordinate axes and passing through the point of intersection of the lines 2x – 5y + 1 = 0 and x – 3y – 4 = 0. [Mar. ’06, ’00]
Solution:
Given the equation of the lines are
2x – 5y + 1 = 0 …….(1)
x – 3y – 4 = 0 …….(2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q13
\(\frac{x}{23}=\frac{y}{9}=\frac{1}{-1}\)
\(\frac{x}{23}\) = -1; \(\frac{y}{9}\) = -1
x = -23; y = -9
∴ Point of intersection of lines (1) & (2) is, P = (-23, -9)
The equation of the straight line in the intercept form is, \(\frac{x}{a}+\frac{y}{b}=1\) = 1 ……..(3)
The straight line (3) makes equal intercepts on the coordinate axes
from (3),
\(\frac{x}{a}+\frac{y}{a}\)
x + y = a ……..(4)
Since equation (4) passes through the point P(-23, -9) then,
-23 – 9 = a
a = -32
∴ The equation of the straight line is x + y = -32
x + y + 32 = 0

Question 14.
Find the length of the perpendicular drawn from the point of intersection of the lines 3x + 2y + 4 = 0 and 2x + 5y – 1 = 0 to the straight line 7x + 24y – 15 = 0. [May ’01; Mar. ’91]
Solution:
Given, the equation of the straight lines are
3x + 2y + 4 = 0 ……(1)
2x + 5y – 1 = 0 ……..(2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q14
∴ The point of intersection of lines (1) & (2) is, P = (-2, 1)
Given the equation of the straight line is 7x + 24y – 15 = 0
Comparing with ax + by + c = 0 then a = 7, b = 24, c = -15
Point of intersection P(x1, y1) = (-2, 1)
The perpendicular distance from P(-2, 1) to the straight line
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q14.1

Question 15.
If θ is the angle between the lines \(\frac{x}{a}+\frac{y}{b}=1\) and \(\frac{x}{b}+\frac{y}{a}=1\), find the value of sin θ, where a > b. [May ’09]
Solution:
Given, the equation of the straight lines are
\(\frac{x}{a}+\frac{y}{b}=1\)
\(\frac{b x+a y}{a b}\) = 1
bx + ay = ab
bx + ay – ab = 0 ……..(1)
\(\frac{x}{b}+\frac{y}{a}=1\)
ax + by = ab
ax + by – ab = 0 ………(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = b, b1 = a, c1 = -ab
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = a, b2 = b, c2 = -ab
If ‘θ’ is the angle between lines (1) & (2) then,
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q15

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 16.
Find the equations of the straight lines passing through (1, 3) and (i) parallel to (ii) perpendicular to the line passing through the points (3, -5) and (-6, 1). [May ’15 (AP)]
Solution:
The slope of the line passing through the points (3, -5) and (-6, 1) is
m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{1+5}{-6-3}=\frac{-6}{9}=\frac{-2}{3}\)
(i) Equation of the line passing through (1, 3) and parallel to the line passing through the points (3, -5) and (-6, 1) is y – y1 = m(x – x1)
⇒ y – 3 = \(\frac{-2}{3}\) (x – 1)
⇒ 3y – 9 = -2x + 2
⇒ 2x + 3y – 11 = 0
(ii) Equation of the line passing through (1, 3) and perpendicular to the line passing through the points (3, -5) and (-6, 1) is y – y1 = \(\frac{-1}{m}\) (x – x1)
⇒ y – 3 = \(\frac{1}{2}\) (x – 1)
⇒ 2y – 6 = 3x – 3
⇒ 3x – 2y + 3 = 0

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 7 Trigonometric Equations to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 1.
Solve 2cos2θ – \(\sqrt{3}\) sin θ+1 = 0
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 1

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 2.
Solve 1 +sin2θ = 3 sinθ cosθ
Solution:
Dividing by cos2 θ we get
⇒ sec2 θ + tan2 θ = 3 tanθ
⇒ 1 + 2 tan2 θ = 3 tanθ
⇒ 2tan2 θ – 3tan θ +1 = θ
⇒ 2 tan2 θ –  2tan θ –  tanθ + 1 = θ
⇒ (tanθ –  1)(2tan θ – 1) = θ
⇒ tanθ = 1 or tanθ = \(\frac{1}{2}\)
If tan θ = 1 then principal solution is α =\( \frac{\pi}{4}\)
General solution is θ = nπ + \(\frac{\pi}{4}\)  n ∈ Z
Let a be the principal solution of tan θ = \(\frac{1}{2}\)
Then the general solution is
θ = nπ+α,n∈ Z

Question 3.
Solve \(\sqrt{2}(\sin x+\cos x)=\sqrt{3}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 3

Question 4.
Solve tanθ+3 cotθ = 5 secθ
Solution:
The given equation is
tanθ + 3 cotθ = 5 secθ
⇒ \(\frac{\sin \theta}{\cos \theta}+\frac{3 \cos \theta}{\sin \theta}=\frac{5}{\cos \theta}\)
⇒  sin2 θ + 3cos2θ= 5 sinθ
⇒  sin2 θ + 3(1 – sin2 θ) = 5 sinθ
⇒ sin2 θ + 3 – 3 sin2 θ = 5 sinθ
⇒ -2 sin θ – 5 sinθ + 3 = 0
⇒ 2 sin θ+ 5 sinθ – 3 = θ
⇒ 2 sin θ + 6 sinθ – sinθ – 3 = θ
⇒ 2 sin θ(sin θ + 3) –  1(sinθ + 3) = θ
⇒ (2sinθ – 1)(sinθ + 3)= θ
⇒ sin θ = \(\frac{1}{2}\) or sinθ = – 3.
sinθ =-3 is not admissible but for sinθ = \(\frac{1}{2}\) general solution is
θ=nπ+(-n)n ,n∈Z
Since the principal solution is α = \(\frac{1}{2}\)

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 5.
If acos2θ + bsin2θ =c has θ12 as its solutions then show that tan θ1 + tan θ1 = \(\frac{2 b}{c+a}\)
Solution:
Given equation is acos2θ + bsin2θ = c
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 4
This is a quadratic equation in an θ.
Given θ12  are the solutions of θ.
tan θ1, tanθ2 are the roots of equation (1)
∴ Sum of the roots tan θ1 + tan θ2 = \(\frac{2 b}{a+c}\) and product of the roots
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 5

Question 6.
Find all values of x in (-π, π) satisfying the equation g1+cosx+cos2 x +……………….. = 43
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 6

Question 7.
Solve sin x = \(\frac{1}{\sqrt{2}}\)
Solution:
Principal solution is α = \(\frac{\pi}{4}\)
General solution is
\(\mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}\)

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 8.
Solve sin 2θ \(=\frac{\sqrt{5}-1}{4}\)
Solution:
The principal solution is α = 18°
∴ General solution is
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 7
Question 9.
Solve tan2 θ = 3.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 8
Question 10.
Solve 3cosec x = 4sinx.
Solution:
Given 3 cosec x = 4 sin x
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 9

Question 11.
If x is acute and sin(x+10°) = cos(3x – 68°) find x in degrees.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 10
When k = 0 we get x = 37°
If we take k = 1, 2 the value of x is not acute.
Hence the value of x is 37°.

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 12.
Solve cos 3θ = sin 2θ.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 11

Question 13.
Solve 7 sin2θ+3cos2 θ= 4.
Solution:
Given 7 sin2θ+3cos2 θ= 4.
⇒ 7 sin 2θ + 3(1 -sin2 θ) = 4
⇒  4 sin 2θ = 1
⇒ sinθ = ± \(\frac{1}{2}\)
Principal solutions are \(\alpha=\pm \frac{\pi}{6}\) and general solution is given by general solution is given by
θ = nπ ± ,\(\alpha=\pm \frac{\pi}{6}\),n∈Z

Question 14.
Find general solution of θ which satisfies both the equations sinθ = \(-\frac{1}{2}\) and cosθ = \(-\frac{\sqrt{3}}{2}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 12

Question 15.
Solve 4 sin x. sin 2x sin 4x = sin 3x
Solution:
Given sin 3x = 4 sinx sin 2x sin 4x
= 2sinx (2sin2x – sin4x)
⇒ 2 sin x (cos 2x – cos 6x)
⇒ sin3x = 2cos 2x sinx – 2 cos 6x sin x
⇒ sin 3x = sin 3x – sin x – 2 cos 6xsinx
⇒ 2cos 6x sinx + sinx= 0
⇒ sinx(2 cos 6x + 1)=0
⇒ sinx = 0 or cos 6x \(-\frac{1}{2}\)

Case (i): sin x = 0,  x= nπ, n∈Z is the general solution.
Case (ii) : cos 6x = \(-\frac{1}{2}\)
Principal solution is α \( =\frac{2 \pi}{3}\)
∴ General solution is 6x = 2πr ± \(\frac{2 \pi}{3}\)
\(x=\frac{n \pi}{3} \pm \frac{\pi}{9}, n \in Z\)

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 16.
If 0<θ<π solve cosθ cos2θ cos3θ = \(\frac{1}{4}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 12
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 13

Question 18.
Solve sin2x – cos2x = sinx – cosx
Solution:
The given equation can be written as
sin 2x – sin x = cos 2x – cos x
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 14

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 19.
Solve cos 3x + cos 2x = sin + sin = \(\sin \frac{3 x}{2}+\sin \frac{x}{2} 0 \leq x \leq 2 \pi\)
Solution:
Given cos 3x + cos 2x = sin + sin = \(\sin \frac{3 x}{2}+\sin \frac{x}{2} 0 \leq x \leq 2 \pi\)
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 15
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 16

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 8 Inverse Trigonometric Functions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions

Question 1.
Prove that
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 14
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 1

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 2.
Find the values of the following.

(i) \(\sin ^{-1}\left(-\frac{1}{2}\right)\)
Solution:
\(\sin ^{-1}\left(-\frac{1}{2}\right)=-\sin ^{-1}\left(\frac{1}{2}\right)=-\frac{\pi}{6}\)

(ii) \(\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 2

(iii) \(\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Solution:
\(\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\tan ^{-1}\left(\tan \frac{\pi}{6}\right)=\frac{\pi}{6}\)

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

(iv) cot-1 (-1)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 3

(v) sec -1  \((-\sqrt{2})\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 4

(vi) Cosec -1  \(\left(\frac{2}{\sqrt{3}}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 15

Question 3.
Find the values of the following.

(i) sin-1 \(\left(\sin \frac{4 \pi}{3}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 6

(ii) \(\tan ^{-1}\left(\tan \frac{4 \pi}{3}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 7

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 4.
Find the values of the following.

(i) \(\sin \left(\cos ^{-1} \frac{5}{13}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 8

(ii) \(\tan \left(\sec ^{-1} \frac{25}{7}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 9

(iii) \(\cos \left(\tan ^{-1} \frac{24}{7}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 10

Question 5.
Find the values of the following.

(i) \(\sin ^2\left(\tan ^{-1} \frac{3}{4}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 11

(ii) \(\sin \left(\frac{\pi}{2}-\sin ^{-1}\left(-\frac{4}{5}\right)\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 12

(iii) \(\cos \left(\cos ^{-1}\left(-\frac{2}{3}\right)-\sin ^{-1}\left(\frac{2}{3}\right)\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 13

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

(iv) sec2 (cot-1 3) + cosec2 (tan-1 2)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 16

Question 6.
Find the value of \(\cot ^{-1}\left(\frac{1}{2}\right)+\cot ^{-1}\left(\frac{1}{3}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 17

Question 7.
Prove that
\(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{7}{25}\right)=\sin ^{-1}\left(\frac{117}{125}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 18

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 8.
If x ∈(-1, 1) prove that 2 tan-1 x = \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)
Solution:
Given x ∈ (-1,1) and it tan-1 x = a then tan α = x and
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 19

Question 9.
Prove that \(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right) +\sin ^{-1}\left(\frac{16}{65}\right)=\frac{\pi}{2}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 20

Question 10.
Prove that cot-1 9+ cosec-1 \( \frac{\sqrt{41}}{4}=\frac{\pi}{4}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 21

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 11.
Show that cot \(\begin{aligned} \cot \left(\operatorname{Sin}^{-1} \sqrt{\frac{13}{17}}\right) \\ = \sin \left(\operatorname{Tan}^{-1}\left(\frac{2}{3}\right)\right) \end{aligned}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 22

Question 12.
Find the value of \(tan \left[2 \operatorname{Tan}^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 23

Question 13.
Prove that \(\operatorname{Sin}^{-1}\left(\frac{4}{5}\right)+2 \operatorname{Tan}^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 24

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 14.
If sin-1 x + sin-1 y + sin-1 z = π, then prove that x4 + y4 + z4 + 4x2y2z2 = 2 (x2y2 + y2z2 + z2x2)
Solution:
Let sin-1 x = A, sin-1 y = B and sin-1z = C
then A+B+C = π …………………..(1)
and sinA = x, sin B = y, sin C = z
Now A+B = π – c
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 25

Question 15.
If \(\operatorname{Cos}^{-1}\left(\frac{p}{a}\right)+\operatorname{Cos}^{-1}\left(\frac{q}{b}\right)\) =α the prove that
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 36
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 26
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 27

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 16.
Solve \(\sin ^{-1}\left(\frac{5}{x}\right)+\sin ^{-1}\left(\frac{12}{x}\right)=\frac{\pi}{2},(x>0)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 28

Question 17.
Solve
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 35
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 30

Question 18.
Solve \(\operatorname{Sin}^{-1} x+\operatorname{Sin}^{-1} 2 x=\frac{\pi}{3}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 31
when \(x=-\frac{\sqrt{3}}{2 \sqrt{7}}\) value is not admissible
Since sin-1 x and sin-1 2x are negative
Hence \(x=-\frac{\sqrt{3}}{2 \sqrt{7}}\)

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 19.
If sin [2 Cos-1 (cot (2 Tan-1x)}] = 0 find x.
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 32

Question 20.
Prove that
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 34
Solution:
Let cot-1 x=θ then cot θ = x and θ <x<π
∴ sin (cot-1x) = sinθ = \(\frac{1}{\operatorname{cosec} \theta}\)
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 33

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 21.
Show that sec2 (tan-1) + cosec2 (cot-1 2) = 10.
Solution:
[1 + tan2 (tan-1(2)] + [1+ cot2 (cot-1(2))]
= 1 + 4 + 1 + 4 = 10

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Telangana TSBIE TS Inter 1st Year Commerce Study Material 2nd Lesson Business Activities Textbook Questions and Answers.

TS Inter 1st Year Commerce Study Material 2nd Lesson Business Activities

Long Answer Questions

Question 1.
What is meant by industry? Explain various types of industries with suitable examples.
Answer:
Industry is concerned with the making or manufacturing of goods. It is that part of the production which is involved in changing the form of goods at any stage from raw material to the finished product. E.g.: Weaving woollen yam into cloth. Thus industry imparts form utility in goods.

Classification or types of industries: The industries may be classified as follows.
1) Primary industry: Primary industry is concerned with production of goods with the help of nature. It is nature-oriented industry, which requires very little human effort. E.g: Agriculture, Farming, Fishing, Horticulture etc.

2) Genetic industry: Genetic industry is related to the reproducing and multiplying of certain species of animals and plants with the object of earning profits from their sale. E.g: Nurseries, cattle breeding poultry, fish hatcheries etc.

3) Extractive industry: It is engaged in raising some form of wealth from the soil, cli-mate, air, water or from beneath the surface of the earth. Generally the products of extractive industries comes in raw farm and they are used by manufacturing and construction industries for producing finished products. E.g: Mining, coal, mineral, iron ore, oil industry, extraction of timber and rubber from forests.

4) Construction industry: The industry is engaged in the creation of infrastructure for the smooth development of the economy. It is concerned with the construction, erection or fabrication of products. These industries are engaged in the construction of buildings, roads, dams, bridges and canals.

5) Manufacturing industry: This industry is engaged in the conversion of raw material into semifinished or finished goods. This industry creates form utility in goods by making them suitable for human uses. E.g: Cement industry, Sugar industry, Cotton textile industry, Iron and steel industry, Fertiliser industry etc.

6) Service industry: In modern times, service sector plays an important role in the development of the nation and therefore it is named as service industry. These are engaged in the provision of essential services to the community. E.g: Banking, trans-port, insurance etc.

Question 2.
What is commerce? Describe its branches.
Answer:
Commerce is concerned with exchange of goods. It includes all those activities which are related to transfer of goods from the places of production to the ultimate consumer. Commerce embraces all those processes which help to break the barriers between producers and consumers. It is the sum total of those processes which are engaged in the removal of hindrances of persons, place, time and exchange.

Branches of commerce:
The activities of commerce may be classified into following two broad categories.

  • Trade
  • Aids to trade

I) Trade: Trade is a branch of commerce. It connects with buying and selling of goods and services. An individual who does trade is called a trader. Trader transfers the goods from the producer to the consumer.

Trade may be classified into a) Home trade, b) Foreign trade. Home trade may again sub-divided into (i) Wholesale trade and (ii) Retail trade.

a) Home trade: Home trade is also known as ‘domestic trade’ or ‘Internal trade’ Home trade is carried on within the boundaries of a nation. Home trade again is of two types: wholesale trade and retail trade.

  • Wholesale trade: It implies buying and selling of goods in large quantities. Traders who engage themselves in whole sale trade are called ‘wholesalers’. Wholesale serves as a connecting link between the producers and the retailers.
  • Retail trade: It involves buying and selling of goods in small quantities. Traders engaged in retail trade are called ‘retailers’. They serve as a connecting link between wholesalers and consumers.

b) Foreign trade: It refers to buying and selling of goods and services between two or more countries through international air ports and sea ports. Foreign trade is also known as ‘external trade’ or ‘International trade’.

  • Export trade: It means the sale of goods to foreign countries. For example India exports tea to the U.K.
  • Import trade: It refers to the purchase of goods from foreign countries. For instance India buys petrol from Iran.
  • Entrepot trade: Importing (buying) goods from one country for the purpose of ex-porting (selling) them to another country is called entrepot trade. This type of trade is also known as re-export trade.

II) Aids to trade: Trade or exchange of goods involves several difficulties, which can be removed by auxiliaries are known as aids to trade. It refers to all those activities, which directly or indirectly facilitate smooth exchange of goods and services.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Aids to trade includes transport, Communication, Warehousing, Banking, Insurance, Advertising. These ensure smooth flow of goods from producers to the consumers. The various aids to trade in commerce are explained in below:

1) Transport: There will be a vast distance between centers of production and centres of consumption. This difficulty is removed by transport. Transport creates place utility. There are several kinds of transport such as air, water and land transport. The geo-graphical distance between producers and consumers is removed with the help of following means of transport.

2) Communication: Communication means transmitting or exchange of information froth one person to another. It can be oral or in writing. It is necessary to communicate information from one to another. Modern means of communication like telephone, email, video conference etc play an important role in establishing contact between businessmen, producers and consumers.

3) Warehousing: There is a time gap between production and consumption. It becomes necessary to make arrangements for storage or warehousing. The goods such as umbrellas and woolen clothes are produced throughout the year but are demanded only during particular seasons like rainy and winter season. Therefore goods need to be stored in warehouses till they are demanded. So, it creates time utility by supplying the goods at right time to consumer.

4) Insurance: Insurance reduces the problem of risks. Business is subject to risks and uncertainities. Risks may be due to fire, theft, accident or any other natural calamity. Insurance companies who act as risk bearer cover risks. Insurance tries to reduce risks by spreading them out over a larger number of people by encouraging them to take insurance policies.

5) Banking: Banking solves the problem of finance. Banking and financial institutions solves the problem of payment and facilitate exchange between buyer and seller. Banks provide such finance to them. Banks also advance loans in the form of overdraft, cash credit and discounting of bills of exchange.

6) Advertising: Advertising fills the knowledge gap. Exchange of goods and services is possible only if producers can bring the products to the consumers. Advertising and publicity are important medias of mass communication. Advertising helps the consumers to know about the various brands manufactured by several manufacturers. The medias used to advertise products are Radio, Newspapers, Magazines, TV, Internet etc.

Question 3.
Discuss the significance of commerce in the present scenario.
Answer:
Commerce can also be defined as The sum total of those processes. Which are engaged in the removal of hindrances of person, place and time in the exchange of commodities’.

Importance of commerce:
The importance of commerce is explained below:
1) Commerce tries to satisfy increasing human wants: Human wants and desires are never ending. Commerce has made distribution and movement of goods possible from one part of the world to the other. Today we can buy anything produced anywhere in the world. Hence commerce facilitates the people to satisfy their needs, desires and wants by distribution and exchange of the goods and services.

2) Commerce helps us to increase our standard of living: Standard of living refers to quality of life enjoyed by the members of a society. When a man consumes more products his standard of living improves. Commerce helps us to get what we want at the right time, right place and at the right price and thus helps in improving our standard of living.

3) Commerce links producers and consumers: Commerce makes a link between producers and consumers through retailers and wholesalers and also through the aids to trade. Thus it creates and facilitates the contact between the centres of production and consumption.

4) Commerce generates employment opportunities: The growth of commerce and trade cause the growth of agencies of trade such as banking, transport, warehousing, insurance, advertising etc. Thus, development of commerce generates more and more employment opportunities.

5) Commerce increases national income and wealth: When production increases, national income also increases. It also helps to earn foreign exchange by way of exports and duties levied on imports.

6) Commerce helps in expansion of aids-to-trade: With the growth in trade and commerce there is a growing need for expansion and modernization of aids to trade. Aids to trade such as banking communication, advertising and publicity, transport, insurance etc. are expanded and modernised for the smooth conduct of commerce.

7) Commerce encourages international trade: With the help of transport and communication development, countries can exchange their surplus. Commodities and earn foreign exchange. Thus, the commerce ensures faster economic growth of the country.

8) Commerce benefit underdeveloped countries: Underdeveloped countries can import skilled labour and technical knowhow from developed countries, while the advanced countries can import raw materials from underdeveloped countries. This helps in laying down the seeds of industrialisation in the underdeveloped countries.

9) Commerce helps during emergencies: During emergencies like floods, earthquakes and wars, commerce helps in reaching the essential requirements like food stuff, medicines and relief measures to the affected area.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Question 4.
Define trade and explain various types of aids to trade.
Answer:
Trade: Buying and selling of goods and services to earn profit is called trade. The person who undertakes this job is known as trader. It is a branch of commerce.

Aids to trade:
Trade or exchange / distribution of goods involves several difficulties, which can be removed by auxiliaries are known as aids to trade. It refers to all those activities which directly or indirectly facilitate smooth exchange of goods and services.

Aids to trade includes Transport, Communication, Warehousing, Banking, Insurance, Advertising. These ensure smooth flow of goods from producers to the consumers. The various aids to trade in commerce are explained in following points.

1) Transport: There will be a vast distance between centres of production and centres of consumption. This difficulty is removed by transport. Transport creates place utility. There are several kinds of transport such as air, water and land transport. The geo-graphical distance between producers and consumers is removed with the help of following means of transport, (i) Land (ii) Water (iii) Air.

2) Communication: Communication means transmitting or exchange of information from one person to another. It can be oral or in writing. It is necessary to communicate information from one to another. Modem means of communication like telephone, email, teleconference, video-conference etc. play an important role in establishing contact between businessmen, producers and consumers.

3) Warehousing: There is a time gap between production & consumption. It becomes necessary to make arrangements for storage or warehousing. The goods such as umbrellas and woolen clothes are produced throughout the year but are demanded only during particular seasons like rainy and winter season. Therefore goods need to be stored in warehouses till they are demanded. So it creates time utility by supplying the goods at right time to consumers.

4) Insurance: Insurance reduces the problems of risks. Business is subject to risks and uncertainities. Risks may be due to fire, theft, accident or any other natural calamity. Insurance companies who act as risk bearer cover risks. Insurance tries to reduce risks by spreading them out over a large number of people by encouraging them to take insurance policies.

5) Banking: Banking solves the problem of finance. Banking and financial institutions solves the problem of payment and facilitate exchange between buyer and seller. Banks provide such finance to them. Banks also advance loans in the form of overdraft, cash credit and discounting of bills of exchange.

6) Advertising: Exchange of goods and services is possible only if producers can bring the products to the consumers. Advertising and publicity are important medias of mass communication. Advertising helps the consumers to know about the various brands manufactured by several manufacturers. The medias used to advertise products are Radio, Newspaper, Magazines, TV; Internet etc.

Question 5.
Explain inter-relationship between industry, commerce and trade.
Answer:
Business is divided into two categories: Industry and commerce. Commerce is again sub-divided into trade and aids to trade practically all of them are closely related to each other. They are inseparable. All of them are parts of the whole business system. Industry and commerce are closely related to each other. Industry cannot exist without commerce and commerce cannot exist without industry because every producer has to find his market for his product to sell. But the producer has no direct connection with the buyer or consumers. Hence, industry needs commerce.

BUSINESS = INDUSTRY + COMMERCE
Commerce is concerned with the sale, transfer or exchange of goods and services. Hence commerce needs industry for the production of goods and services. Commerce makes the necessary arrangement for linking between producers and ultimate consumers. It includes all those activities that are involved in buying, selling, transporting, banking, warehousing of goods, and insurance of safeguarding the goods.

COMMERCE = TRADE + AIDS TO TRADE
Thus industry, commerce, and trade are closely related to one another and are inter-dependent as shown in the figure below. In conclusion, we can say that industry, trade and commerce are inter-related with each other. Industry is concered with production of goods and services and commerce arranges its sales; but the actual operation of sales is in the hands of trade. So they cannot work independently.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Question 6.
Describe the various Hindrances of Commerce.
Answer:
Commerce is concerned with exchange of goods. It includes all those activities which are related to transfer of goods from the place of production to the ultimate consumers. Whereas trade involves buying and selling of goods, commerce has a wider meaning. Commerce include trade and aids to trade. The aids to trade include transport, banking, insurance, warehousing, advertisement and salesmanship.

Hindrances of trade: In the course of exchange of goods various problems are encountered. The hindrances in the way of smooth trade may be place, person, finance, time, knowledge and risk.

1) Hindrances of place: Generally, all the goods are not consumed at the same place where are produced. The goods are to be taken from a place where there is less demand, to the place where there is more demand. The activity of movement of the goods is called transportation. Thus transport eliminates the hindrances of place.

2) Hindrances of persons: In the present day world the consumers are in millions and it is not possible for the producers to know the consumers who are in need of goods produced by them. A chain of middlemen like wholesalers, retailers, dealers etc. Purchase goods from the producers and take them to the customers. Thus, middlemen remove the hindrances of persons.

3) Hindrances of finance: There is always time lag between the production and sale of goods. It takes time to collect money and hence need finance for trade. Commerce makes exchange of goods and services possible by removing these hindrances through the agency of banks.

4) Hindrances of time: As the goods are produced in anticipation of demand, there is a need to store the until they are required for consumption. Warehousing eliminates the hindrances of time and provides time utility to goods.

5) Hindrances of knowledge: The consumers may not be aware of the availability of various goods in the market. The absence of information is another hindrance. This is eliminated through advertising. Advertisement is done through T.V., radio, news papers, magazines, wall posters, hoardings etc.

6) Hindrances of risk: There are risks involved in production, transporting goods from one place to another, warehousing. The businessmen would like to cover these risks. Insurance companies undertake to compensate the loss suffered due to such risks. So, insurances eliminates hindrances of risks.

Question 7.
Distinguish between trade, commerce and industry.
Answer:
Differences between trade, commerce and industry.

Basis of Difference Trade Commerce Industry
1. Meaning It is related to the purchase and sale of goods. It is related to the activities which deals with taking of goods from producers to consumers. All those activities which deal with conversion of raw material into finished goods.
2. Capital More capital is required than commerce. It requires less capital. Capital needs are high.
3. Scope It deals with purchase and sale of goods. It includes trade and aids to trade. Primary manufacturing, processing etc., industries are covered under industry.
4. Risk It involves greater risk. The risk involved is comparatively less. It involves greater risk compared to any other activity.
5. Utility Trade creates possession utility. Commerce creates time and place utility. Industry creates form utility.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Short Answer Questions

Question 1.
Define Industry.
Answer:
Industry is concerned with the making or manufacturing of goods. It is that part of the production which is involved in changing the form of goods at any stage from raw material to the finished product. E.g.: Weaving woollen yam into cloth. Thus industry imparts form utility in goods.

Question 2.
What is Trade?
Answer:
Trade: Buying and selling of goods and services to earn profit is called trade. The person who undertakes this job is known as trader. It is a branch of commerce.

Trade is a branch of commerce. It connects with buying and selling of goods and services. An individual who does trade is called a trader. Trader transfers the goods from the producer to the consumer.

Trade is classified into two types: (I) Home trade (II) Foreign trade

I) Home trade: Home trade is also known as ‘domestic trade’ or ‘Internal trade’. Home trade is carried on within the boundaries of a nation.
Home trade is again divided into two types:

  • Wholesale trade and
  • Retail trade

(i) Wholesale trade: It implies buying and selling of goods in large quantities. Traders who engage themselves in wholesale trade are called ‘wholesalers’. Wholesale serves as a connecting link between the producers and the retailers.

(ii) Retail trade: It involves buying and selling of goods in small quantities. Traders en-gaged in retail trade are called ‘retailers’. They serve as a connecting link between wholesalers and consumers.

(II) Foreign trade: It refers to buying and selling of goods and services between two or more countries through international air ports and sea ports. Foreign trade is also known as ‘external trade’ or ‘international trade’. Foreign trade is again may be classified into three categories as mentioned below.

  • Export trade: It means the sale of goods to foreign countries. For example India exports tea to the united kingdom.
  • Import trade: It refers to the purchase of goods from foreign countries. For instance India buys petrol from Iran.
  • Entrepot trade: Importing (buying) goods from one country for the purpose of ex-porting (selling) them to another country is called entrepot trade. This type of trade is also known as ‘re-export’ trade.

Question 3.
State the types of foreign trade.
Answer:
When trade takes place between two countries it is called foreign trade or external trade or international trade. Two countries are involved in foreign trade. External trade generally requires permission from the respective countries. The hindrances of place, time, risk, exchange are overcome with the help of various agencies. Foreign trade may be classified into export trade, import trade and entrepot trade.

1) Export trade: This trade refers to sale of goods to foreign countries.
E.g.: India exports tea to U.K.

2) Import trade: The purchase of goods from other countries is known as import trade.
Ex: India buys petrol from Iran.

3) Entrepot trade: When goods are imported (purchased) from one country with a view to exporting (selling) them to other country, it is called entrepot trade or re-export trade.
Ex: India imported petrol from Iran and export the same to Nepal.

Question 4.
Explain the classification of industries.
Answer:
Classification or types of industries: The industries may be classified as follows.
1) Primary industry: Primary industry is concerned with production of goods with the help of nature. It is nature-oriented industry, which requires very little human effort.
E.g: Agriculture, Farming, Fishing, Horticulture etc.

2) Genetic industry: Genetic industry is related to the reproducing and multiplying of certain species of animals and plants with the object of earning profits from their sale.
E.g: Nurseries, cattle breeding poultry, fish hatcheries etc.

3) Extractive industry: It is engaged in raising some form of wealth from the soil, cli-mate, air, water or from beneath the surface of the earth. Generally the products of extractive industries comes in raw farm and they are used by manufacturing and construction industries for producing finished products.
E.g: Mining, coal, mineral, iron ore, oil industry, extraction of timber and rubber from forests.

4) Construction industry: The industry is engaged in the creation of infrastructure for the smooth development of the economy. It is concerned with the construction, erection or fabrication of products. These industries are engaged in the construction of buildings, roads, dams, bridges and canals.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

5) Manufacturing industry: This industry is engaged in the conversion of raw material into semifinished or finished goods. This industry creates form utility in goods by making them suitable for human uses.
E.g: Cement industry, Sugar industry, Cotton textile industry, Iron and Steel industry, Fertiliser industry etc.

6) Service industry: In modern times, service sector plays an important role in the development of the nation and therefore it is named as service industry. These are engaged in the provision of essential services to the community.
E.g: Banking, Trans-port, Insurance etc.

Question 5.
Define ‘Entrepot trade’ with example.
Answer:
When goods are imported from one country and the same are exported to another country, such trade is called entrepot trade.
E.g.:

  • India importing wheat from U.S. and exporting the same to Srilanka.
  • India imports petrol from Iran and export the same to Nepal.

Very Short Answer Questions

Question 1.
Industry.
Answer:
Industry is concerned with the making or manufacturing of goods. It is that part of the production which is involved in changing the form of goods from raw material to the finished product.
E.g.: Weaving woollen yam into cloth. Thus industry creates form utility in goods.

Question 2.
Commerce.
Answer:

  • Commerce is concerned with exchange of goods. It includes all those activities which are related to transfer of goods from the places of production to the ultimate consumer. Commerce embraces all those processes which help to break the barriers between producers and consumers.
  • It is the sum total of those processes which are engaged in the removal of hindrances of persons, place, time and exchange.

Question 3.
Trade.
Answer:

  • All the activities engaged in buying and selling of goods and services a re called trade.
  • Therefore trade includes sale, transfer or exchange of goods and services with the intention of making profit. The object of trade is to make goods available to those who need them and willing to pay for them.
  • Trade is the final stage of business activities and involves transfer of ownership.

Question 4.
Home Trade.
Answer:

  • The trade is carried on within the boundaries of the nation is called Home Trade. Home trade is also called as internal trade or domestic trade.
  • Home trade refers to a trade where buying and selling of goods takes place between the persons who belong to the same country. So, home For movement of goods internal transport system is used.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Question 5.
Entrepot trade.
Answer:

  • When goods are imported from one country and the same are exported to another country, such trade is called entrepot trade.
  • E.g.: India importing wheat from U.S. and exporting the same to Srilanka.

Question 6.
Transportation.
Answer:
There is a vast distance between centres of production and centres of consumption. Goods are to be moved from the places of production to the place where they are demanded.

  • The activity which is concerned with movement of goods is called transportation. Trans-port create place utility.
  • There are several kinds of transport such as air, water and land transport. The geo-graphical distance between producers and consumers is removed with the help of transport.

Question 7.
Warehousing.
Answer:

  • There is time gap between production and consumption. Hence, it became necessary to make arrangements for storage or warehousing.
  • It is one of the aid to trade, which facilitates to store the goods until they get demand or consumed.
  • Some goods are to be stored in warehouses till they are demanded. Warehousing creates time utility.

Question 8.
Genetic Industries.
Answer:

  • Genetic industry is related to the reproducing and multiplying certain species of animals and plants with the object of earning profit from their sale.
  • E.g.: Nurseries, cattle breeding, poultry farm, fish hatcheries etc.

Question 9.
Extractive industries.
Answer:

  • These industries are engaged in raising some form of wealth from the soil, climate, air, water or from beneath the surface of the earth. Generally the products of extractive industry comes in raw form and they are used by manufacturing and construction industries for producing finished products.
  • E.g.: Mining, coal, mineral, iron ore oil, extraction of timber and rubber from forests.

Question 10.
Banking.
Answer:

  • Banking solve the problem of finance.
  • Producers and traders require money for carrying on production and trade. Banks are the institutions which supply funds for industries and trade.
  • They pool savings from the public and make them available to industry. So, banking is an important function of commerce.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Telangana TSBIE TS Inter 1st Year Commerce Study Material 1st Lesson Introduction to Business Textbook Questions and Answers.

TS Inter 1st Year Commerce Study Material 1st Lesson Introduction to Business

Long Answer Questions

Question 1.
Define Business. Explain its characteristic features.
Answer:
Meaning:

  • Business means the state of being busy.
  • Business is an economic activity involving production, exchange, distribution and sale of goods and services with an objective of making profits and maximization of wealth.
  • The primary intention of business is making profits.

Definitions:

  • L.H. Haney: Defined business as “A human activity directed towards producing or acquiring wealth through buying and selling of goods”.
  • Stephen son: Defines business as, “The regular production or purchase and sale of goods undertaken with an objective of earning profits and acquiring wealth through the satisfaction of human wants”.
  • According to keith and carlo: “Business is a sum of all activities involved in the production and distribution of goods and services for private profits”.

Business may be defined as an economic activity which involves regular transfer or exchange of goods or services for a price.

Characteristics (or) Features:
Following are the main characteristic features of business:
i) Economic Activity: Business is an economic activity. It is performed with the main motive of earning money or profit. It does not include the activities undertaken out of love, affection, charity and religious commitments etc.

ii) Deals with goods and services: Every business enterprise produces or buys goods and services with the intention of selling them to others, to earn profit. Goods deals in business, may be consumer goods or capital goods.

  • Consumer goods are meant for direct use by the ultimate consumers, e.g. cloths, note books, tea, bread, etc.
  • Capital goods also known as producer goods, which are not intended for direct consumption but used for production of other goods, Equipment, Tools, Raw mate-rials, Machinery etc.

Services like transport, warehousing, banking, insurance, etc. may be considered as intangible and invisible goods. Services facilitate buying and selling of goods by over coming various hindrances in trade.

iii) Creation of utilities: Business makes goods more useful to satisfy human wants. It adds, time, place, form and possession utilities to various types of goods.
For example: It carries goods from place of production to the place of consumption (place utility). It makes goods available for use in future through storage (time utility).

iv) Continuity in dealings: Dealings in goods and services become business only if undertaken on a regular basis. A single isolated transaction of purchase and sale does not constitute business.
Therefore, regulartity of dealings is an essential feature of business.

v) Sale, transfer or exchange: All business activities involve transfer or exchange of goods and services for some consideration. The consideration called price, which is expressed in terms of money.
For example: If a person cooks and serves food to his family, it is not business. But when he cooks food and sells it to others for a price, it becomes business.

vi) Profit motive: The primary objective of business is to earn profits. Profits are essential for the survival as well as growth of any business. Profits must be earned through legal and fair means. Business should never exploit any part of the society to make money.

vii) Risk and uncertainity: Risk means fear of loss. It implies the uncertainly of profit or the possibility of loss. Business enterprises function in uncertain and uncontrollable environment.

Changes in customers tastes and fashions, demand, competition, government policies etc., create risk, flood, fire, earthquake, strike by employees, theft etc., also cause loss. A businessman can reduce risks through correct forecasting and insurance.

viii) Business is also a social institution: Business is also a social institution because it helps to improve the living standards of people through effective utilisation of scarce resources of the society.

ix) Art as well as science: Business can be pressumed as an art as well as science. It is art because it requires personal skills and experience. It is also a science because it is based m certain principles and laws.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Question 2.
Describe various objectives of a business.
Answer:
“An objective is a goal or motive directed towards achievement of predetermined purpose or aim. Hence, every business enterprise has certain objectives. These objectives are broadly classified into four categories viz, economic objectives, social objectives, human objectives and national objectives.

I. Economic Objectives:
Business is basically an economic activity. Therefore, its primary objectives are economic in nature. The main economic objectives of business are as follows.
i) Earning profits:

  • Any business enterprise is established for earning some profits. It is the main in-centive and hope to start business.
  • Just as a person cannot live without food, a business firm cannot survive without profits.
  • Profit is also necessary for the expansion and growth of business.
  • Profit also serves as the barometer of economic stability, efficiency and progress of a business enterprise.
  • Therefore, profit is essential for the survival of every business.

ii) Creating customers:

  • A business man can earn profits only when there are enough customers to buy and pay for his goods and services.
  • In the words of peter F. Drucker, “There is only one valid definition of business purpose; to create a customer.
  • The customer is the foundation of business and keeps it in existence.
  • In order to earn profits, business must supply better quality goods and services at reasonable prices. Therefore, creation and satisfaction of customers is an important economic objective of business.

iii) Innovation:

  • Application and adoption of new methods, devices and technologies is called innovation. It also refers to creation of new things resulting from the study and experimentation, research and development.
  • It comprises all efforts made in perfecting the product, minimising the costs and maximizing benefits to customers. Business firms invest money, time and efforts in Research and Development (R&D) to introduce innovation.

iv) Optimum Utilization of Resources:

  • The best usage of resources like material, machine, men and money is called optimum utilization of resources. It is also one of the important economic objectives.
  • Waste control and reprocessing mechanism, providing proper training to its workers and effective spending of money may prompt the business concern to reach the objective of optimum utilization of resources.

II. Social Objectives:
Business does not exist in a vaccum. It is part of society. It cannot survive and grow without the support of society. Business must therefore discharge social responsibilities in addition to earning profits.

According to Henry Ford, ‘The primary aim of business should be service and subsidiary aim should be earning of profits.”

The social objectives of business are as follows:
i) Supplying desired goods and Services at reasonable prices: Supply of goods and services to consumers at reasonable price with the accepted quality is the responsibility of any business entity. It is also the social obligation of business to avoid malpractices like, adulteration, smuggling, black marketing and misleading advertising.

ii) Fair Remuneration to employees: Business must give fair remuneration and compensation to employees for their work. In addition to wages and salary a reasonable part of profits should be distributed among employees by way of bonus. Such sharing of profits will help to increase the motivation and efficiency of employees. It is the obligation of business to provide healthy and safe work environment for employees.

iii) Employment creation: In a country like India unemployment Has become a serious problem and no government can offer jobs to all. Therefore, provision of adequate and full employment opportunities is a significant service to society.

iv) Promotion of social welfare: Business should provide support to social, cultural and religious, organisation. Biteiness enterprise can build schools, college, Libraria, dharmashalas, Hospitals, sports bodies, and research institutions. They can help non-government organisations (NGOs) like CRY (Child Relief and You), Help Age, and other which render services to weaker sections of society.

v) Payment of dues and Taxes to government: Every business enterprise should pay taxes and dues (income tax and GST) to the government honestly and at the right time.

III. Human objectives: Human objectives are concerned with the well being of labour, satisfaction of customers and shareholders. The human objective are as follows.
i) Welfare of workers or staff: The workers should be provided win physical comfort, appreciation and dignity of labour and conditions which will motivate the workers to give their best. Adequate provisions should be made for their health, safety and social security.

ii) Development of human resources: Human resources are the most valuable asset of business and their development will help in the growth of the business. This can be done by training the employees and conducting workshops on skill development and attitude.

iii) Labour participation: The workers should be allowed to take part in decision making process and helps them in their development.

iv) Labour management co-operation: The employees should be looked upon as human beings. The employees help in increasing probability and should be rewarded for their hardwork.

IV. National objectives: The following are the national objectives of the business.
i) Optimum utilisation of resources: Juditions allocation and optimum utilisation of scarce resources is essential for rapid economic growth of the country.

ii) National Self-reliance: It is the duty of the business to help the government, in increasing exports and in reducing dependence on imports. It will help nation to become self-reliant.

iii) Development of small scale industries and MSMEs : Small scale industries are necessary for generating employment opportunities and for providing inputs to large scale industries.

iv) Development of backward areas: In order to achieve balanced regional development, industries should be started at backward regions which helps to raise the standard of living in backward areas.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Question 3.
Discuss the social responsibility of business.
Answer:
Meaning: Every business operates within a society. It uses the resources of the society and depends on the society for its functioning. This creates an obligation on the part of the business to look after the welfare of the society. Therefore, all the activities of the business should be that they will protect and contribute to the interests of the society. Social responsibility of business refers to all such duties and obligations of business directed towards the welfare of the society.

Responsibility towards different interested groups: The business is associated with owners, employees, suppliers, customers, government and society. They are called as interested groups because, every activity of the business will affect the interests directly or indirectly. Various responsibilities of business towards different groups given below:

1) Responsibility towards owners: Owners are the persons who own the business. They contribute the capital and bears the business risks. The responsibilities of business towards them are:

  • Running the business efficiently.
  • Proper utilisation of capital and other resources.
  • Growth and appreciation of capital
  • Regular and fair returns on capital invested by way of dividends.

2) Responsibility towards employees: The future of the business depends on the efforts and efficiency of the employees. So, it is the responsibility of the take care of their interest. The responsibilities are:

  • Regular and fair payment of their wages and salaries.
  • Better working conditions and welfare amenities.
  • To improve the skill and efficiency by providing proper training.
  • Job security and social security like group insurance, pension, retirement benefits etc.

3) Responsibility towards suppliers: Suppliers are those persons who supply raw ma-terials and other items to the business. The responsibilites towards them are:

  • Timely payment of dues.
  • Dealing on fair terms and conditions.
  • Availing reasonable credit period.

4) Responsibility towards customers: In order to survive, the business responsibility is to provide following facilities of customers:

  • Providing qualitative goods and services.
  • Charging reasonable prices.
  • Giving delivery of goods within stipulated time.
  • After sale service.
  • Avoiding unfair means like under weighing the product, adulteration etc.

5) Responsibility towards government: Business is governed by the rules and regulations framed by the government. The responsibilities towards government are:

  • Payment of fees, duties and taxes honestly and regularly.
  • Setting up units as per guidelines of the government.
  • Following the pollution control norms.
  • Not to indulge in unlawful activities.

6. Responsibility towards society: Business, being a part of the society has to maintain relationship with other members of the society. The responsibility of business towards society is:

  • To help weaker and backward sections of the society.
  • To generate employment.
  • To protect environment.
  • To conserve natural resources and wild life.
  • To promote sports, social and cultural values.

Question 4.
Classify and describe each type of Economic activities.
Answer:
All human activities are directed towards satisfying human wants. Depending upon the nature of wants, human activities may be categorised as economic and non-economic. Economic activities are undertaken to create utilities. Non – economic activities do not have economic values and these are primarily tend to satisfy social, religious, cultural or sentimental requirements of human beings.

Classification of economic activities: Economic activities are broadly classified into three. They are:

  • Business
  • Profession
  • Employment

1) Business: The business is an activity which is primarily pursued with the object of earning profit. A business activity involves production, exchange of goods and services to earn profits or to earn a living. The word business literally means a state of being busy. Every person is engaged in some kind of occupation, a farmer works in the field, a worker works in the factory, a clerk does his work in the office, a teacher teaches in the class, a salesman is busy in selling the goods and an entrepreneur is busy in running his factory. The primary aim of all these persons is to earn their live¬lihood while doing some work.

2) Profession: Profession is an occupation involving the provision of personal services of a specialised and expert nature. The services are based on professional service of specialised nature. The service is based on professional education, knowledge, train¬ing etc. The specialised service is provided for a professional fees charged from the clients. For instance, a doctor helps his patients through his expert knowledge of the science of medicine and charges a fees for the service. Minimum educational qualifications are prescribed for entry into a profession and every profession requires a high degree of formal education and specialised training in a particular field. For example, a chartered accountant needs to be a member of the Institute of Chartered Accountants of India (ICA1). A person entering law profession has to acquire LL.B degree in order to become a lawyer.

3) Employment or Service: Employment or service involves working under a contract of employment for or under someone known as employer in return for a wage or salary. The person engaged under employment works as per the directions of the employer. There is an employer – employee relationship. A professional may also work under the contract of employment. A Chartered Accountant may be employed by the company The service may be of government department or in a private organisation.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Short Answer Questions

Question 1.
What are the Business objectives?
Answer:
Business objectives are classified as economic, Social, human and national objectives.
1. Economic objectives:

  • Earning profit
  • Creation of customers
  • Innovation
  • Optimum utilization of resources.

2. Social objectives:

  • Supplying desired goods and services at reasonable prices.
  • Fair remuneration to employees
  • Creation of employment opportunities
  • Promotion of social welfare
  • Payment of dues and taxes to Government.

3. Human objectives:

  • Welfare of workers or staff
  • Development of human resources
  • Labour participation in management
  • Labour management co-operation.

4. National objectives:

  • Optimum utilisation of natural resources
  • Self-reliance.
  • Development of small scale industries and MSMES
  • Development of backward regions.

Question 2.
What are the economic activities.
Answer:
Economic activities are broadly classified into business, profession and employment.

Business: The word business literally means a state of being busy. Every person is engaged in some kind of occupation or other work. The business is an activity which is primarily pursued with the object of earning profit. So, a business activity involves production, exchange of goods and services to earn profits or to earn a living.

Profession: Profession is an occupation involving the provision of personal services of a specialised and expert nature. The service is based on professional education, knowledge, training etc. The specialised service is provided for a professional fees charged from the clients. For example, a doctor helps his patients through his expert knowledge of science of medicine and charges a fees for the service.

Employment: Employment involves working under a contract of employment for or under someone known as employer in return for a salary. The person engaged under employment works as per the directions of the employer.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Question 3.
What are the social objectives?
Answer:
Business does not exist in a vaccum. It is a part of society. It cannot survive and grow without the support of society. Business must therefore discharge social responsibilities in addition to earning profits.

According to Henry Ford, “The primary aim of business should be service and subsidiary aim should be earning of profit”.

The social objectives of business are as follows:
i) Supplying desired goods and services at reasonable prices: Supply of goods and services to consumers at reasonable price with the accepted quality is the responsibility of any business entity. It is also the social obligation of business to avoid malpractices like, adulteration, smuggling, black marketing and misleading advertising.

ii) Fair Remuneration to employees: Business must be given fair remuneration and compensation to employees for their work. In addition to wages and salary a reasonable part of profits should be distributed among employees by way of bonus. Such sharing of profits will help to increase the motivation and efficiency of employees. It is the obligation of business to provide healthy and safe work environment for employees.

iii) Employment Generation: In a country like India unemployment has become a serious problem and no government can offer jobs to all. Therefore, provision of adequate and full employment opportunities is a significant service to society.

iv) Promotion of social welfare: Business should provide support to social, cultural and religious organisations. Business enterprise can build schools, colleges, libraries, dharamashalas, hospitals, sports bodies and research institutions. They can help NGO’s like CRY (Child Relief and You). Help Age, and others which render services to weaker sections of society.

v) Payment of dues and taxes to government: Every business enterprise should pay taxes and dues (income tax and GST) to the government honestly and at the right time.

Question 4.
What are the National objectives?
Answer:
National objectives of business are as follows:
i) Optimum utilisation of resources: Business should use the nation’s resources in the best possible manner. Judicious allocation and optimum utilisation of scarce resources is essential for rapid and balanced economic growth of the country.

ii) National self-reliance: It is the duty of business to help the government in increasing exports and in reducing dependance on imports. This will help a country to achieve economic independence.

iii) Development of MSMEs: Big business firms are expected to encourage growth of micro, small and medium enterprises (MSMEs) which are necessary for generating employment. MSMEs can be developed as ancillaries which provide inputs to large scale industries.

iv) Development of backward classes: Business is expected to give preference to the industrialisation of backward regions of the country. It will also help to raise standard of living in backward areas. Balanced regional development is necessary for peace and progress in the country. Government offers special incentives to the businessmen who setup factories in notified backward areas.

Question 5.
What is the role of profit in business?.
Answer:
Earning profit is necessary not only to pay adequate returns to the investors, but also to maintain the stability of the business. The role of profit in business is given below:

  • Profits ensure adequate funds for future expansion and innovation, there by increasing the wealth of country.
  • Profits are also helpful to attract new capital from outside sources like banks, Financial Institutions and Investors.
  • Profits are serves as the barometer of economic stability, efficiency and progress of a business enterprise.
  • Profits enable a businessman to stay in business by maintaining intact the wealth producing capacity of its resources.
  • Profits ensure continuous flow of capital for the modernisation of business.

Just as a person cannot line without food, a businesss firm cannot survive without profit. Therefore, profits are essential for every business organisation.

Question 6.
Explain the History of commerce.
Answer:
Commerce is a wider system which is the custome of a gradual evolution spread over a long period of human history and it passes through the following different stages.
i) Household economy: This is the first stage of economic development. At this stage, division of labour was unknown concept at the family level. There was no commercial inter relationship between families. While men engaged in jobs like hunting, fishing, making weapons for hunting etc, women were engaged in fruit gathering, cultivation of lands etc. Therefore, commerce was unknown at this stage.

ii) Primitive Barter Economy: Gradually the need of the families increased and families started specializing in different occupations and the need for exchange of goods, etc., between different families. Thus, commerce originated with introduction of barter systesms. In barter system, the goods are exchanged only. Hence barter system was considered as a popular element of commerce.

iii) The raise of trade: In course of time, the needs of people increased. In the begining, goods were exchanged at particular fixed places, but gradually trade appeared on the scene. Home trade began to develop and assume importance and money appeared as an instrument and medium of exchange. Further, the systems of weights and measurement also came into existence.

iv) Town Economy: At this stage, trade began to be undertaken for catering to the needs of local markets which gradually developed into large town. Further, traders were divided into wholesale and retail merchants. Division of labour become significant and prices for goods began to be fixed regularly and traders started using the credit system in their transactions.

v) International Trade: At this stage, goods were produced not only for selling in the local markets but also in foreign markets. This expansion of trade was due to the industrial revolution which made large scale manufacturing of goods possible. Commercial banks, insurance companies, transport companies, warehousing companies etc began to setup.

vi) E-commerce: It is an innovative business idea in the modem economy e-commerce means electronic commerce. It is the process of buy and sell the goods and services through an electronic medium electric commerce emerged in the early 1990s, and its use has increased at a rapid rate.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Question 7.
What are the economic objectives of the business?
Answer:
Economic objectives:
Business is basically an economic activity. Therefore, its primary objectives are economic in nature. The main economic objectives of business are as follows:

i) Earning profits: Any business enterprise is established for earning some profits. It is the main incentive and hope to start business. Just as a person cannot live without food, a business firm cannot survive without profit. Profit also serves as the barometer of economic activity, efficiency and progress of a business enterprise. Therefore profit is essential for the survival of every business.

ii) Creating customers: A business man can earn profits only when there are enough customers to buy and pay for his goods and services. In the words of Peter F. Drucker, ‘There is only one valid definition of business purpose, to create a customer. The customer is the foundation of business and keeps it in existence. In order to earn profit, business must supply better quality, goods and services at reasonable prices. Therefore, creation and satisfaction of customers is an important economic objective of business.

iii) Innovation: Application and Adoption of new methods, devices and technologies is called Innovation. It also refers to creation of new things resulting from the study and experimentation, research and development. It comprises all efforts made in prefecting the product, minimising the costs and maximizing benefits to customers. Business firms invest money, time and efforts in Research and Development (R and D) to introduce innovation.

iv) Optimum Utilization of Resources: The best usage of resources like material, machine men and money is called optimum utilization of resources. It is also one of the important economic objectives waste control and reprocessing mechanism, providing proper training to its workers and effective spending of money may prompt the business concern to reach the objective of optimum utilization of resources.

Question 8.
What is profession and explain features and objectives of profession.
Answer:
Profession: A profession is an occupation or vocation intended to render personal service of a specialised and expert nature. Profession is based on educational qualification, knowledge and competencies. For providing this service, the service provider may be a doctor, an engineer, lawyer, or a chartered accountant will charge fee on their clients which will be their primary earning.

Following are the salient features of profession:

  • A person enters into any particular profession should have Specialized qualification, knowledge and training in that respective profession.
  • Person enters into the specialized profession must be the member of respective professional body.
    Ex: Doctors in IMA (Indian Medical Association), Lawyers in Indian Bar Council and chartered Accountant in ICA (Institute of Charatered Accountants of India).
  • Person choosing the profession must follows the ethics.
  • Professional fee from the clients for providing their specialized services.
  • Rendering service legibly is the key factor in any profession.
  • Service motive must be an integral part of any profession.
  • Some professionals instead of work on their own may also work in organizations as employees and consultants.

Following are the important objectives of any profession:

  • To secure respected position in the society.
  • To spread the efficiency, skill and knowledge possessed, for the needy people.
  • To build career and register growth and development in their respective field.
  • To render service.
  • To enjoy the freedom of work at the cost low risk.

Question 9.
What is employment and explain the features and objectives of employment.
Answer:
Employment: An employment is a contract of service. A person who works under the contract for a salary or wage is called an employee and the person who has given the job to the employee is called employer. An employee works under an agreement as per the rules of service and performs tasks assigned to him by the employer.

The relationship between two parties will be employer – employee. Working in Banks, offices, hotels, schools and companies are the examples of employment.

Following are the important features of employment:

  • Employment contract commences when employee signs the agreement and joins organisation.
  • It involves rendering service for periodical remuneration called wage or salary.
  • Employment is the contractual relationship between employee and employer.
  • Employment will have very low risk and uncertainty, when compared with other economic activities.
  • Employee need not invest any capital.
  • Employees need to follow service rules and regulations framed by the employer strictly.
  • Certain employments required necessary qualifications too.
  • There are certain terms and conditions, of work like hours of work, duration of work, leave facility, salary / wages and place of worts in employment, which must be followed by the both employer and employee.

The important object of employment can be prevented as under:

  • The basic objective of employment is to earn for their livelihood and support family.
  • To get secured and assured wages and salaries.
  • To free from risk and uncertainty.
  • To keep away from disputes, conflicts and devastations.
  • To focus on skill development and career paths for job seekers.

TS Inter 1st Year Commerce Study Material Chapter 1 Introduction to Business

Very Short Answer Questions

Question 1.
Business.
Answer:

  • Business means the state of being busy.
  • Business is an economic activity involving production, exchange, distribution and sale of goods and services with an objective of making profits and maximization of wealth. The primary intension of business is making profits.

Definition: L.H.Maney defined business as “a human activity directed towards producing or acquiring wealth through buying and selling of goods”.

Question 2.
Human activities.
Answer:

  • The activities performed by the human being in their daily life are called human activities. These are undertaken by them to fulfill their needs, desires, wants and luxuries.
  • Human activities are divided into two types:
    • Economic activities
    • Non-economic activities.

Question 3.
Profession.
Answer:

  • A profession is an occupation or vocation intended to render personal service of a specialized and expert nature. It is part of economic activities undertaken by the human being.
  • Always profession is based on educational qualification, knowledge and competencies. A fee is charged for providing professional service.

Question 4.
Employment.
Answer:

  • Employment is the relationship between two parties. One party render this service on contractual basis for the remuneration wage or salary.
  • Hence the relationship between these two parties will be employer – employee.

Question 5.
Risk and uncertainity.
Answer:

  • Risk means fear or loss. It implies the uncertainity of profit or the possibility of loss. Business enterprises function is uncertain and uncontrollable environment.
  • Changes in customers tastes and fashions, demand, competition, Government policies etc., create risk. Flood, fire, earthquake, strike by employees, theft etc., also cause loss. A businessman can reduce risks through correct forecasting and insurance.

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations important questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 6 Trigonometric Ratios up to Transformations to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Important Questions

Question 1.
Prove that \(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=\cot 36^{\circ}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 1

Question 2.
Find the period of the function defined by f (x) = tan (x+ 4x + 9x+…………+ n2x)
Solution:
The given function is
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 2

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 3.
If A is not an integral multiple of \(\frac{\pi}{2}\), prove that
(i) tan A + cot A = 2 cosec 2A
(ii) cot A – tan A = 2 cot 2A
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 3
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 4

Question 4.
If ABC are angle of a triangle then prove that
\(\begin{aligned} \sin ^2 \frac{A}{2} & +\sin ^2 \frac{B}{2}-\sin ^2 \frac{C}{2} \\
& =1-2 \cos \frac{A}{2} \cos \frac{B}{2}-\sin ^2 \frac{C}{2} \end{aligned}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 5

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 5.
If tan 20° = λ then show that
\(\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \tan 110^{\circ}}=\frac{1-\lambda^2}{2 \lambda}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 6

Question 6.
If \(\cos \theta+\sin \theta=\sqrt{2} \cos \theta\) then show that \(\cos \theta-\sin \theta=\sqrt{2} \sin \theta\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 7

Question 7.
Show that cos 340° cos 40° + sin 200° sin 140°=\(\frac{1}{2}\)
Solution:
LH.S = cos 3400 cos 40° + sin 2000 sin 140°
= cos (360 – 20°) cos 40° + sin (180 + 20°) sin (180 – 40°)
= cos 20° cos 40° – sin 20° sin 40°
= cos (20° + 40°) = cos 60° = R.H.S

Question 8.
Find the value of tan 100° + tan 125° + tan 100° + tan 125°
Solution:
We have tan 100° + tan 125° = 225
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 8
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 9

Question 9.
Prove that tan 500° – tan 400° = 2 tan 10°
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 10

Question 10.
Show that cos 42° + cos 78° + cos 162° = 0
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 11

Question 11.
Find the value of \(\cos ^2 52 \frac{1}{2}^{\circ}-\sin ^2 22 \frac{1}{2}^{\circ}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 12

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 12.
If A+B+C= \(\frac{\pi}{2}\) then show that cos A + cos B + cos C
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 13

Question 13.
If tanθ=\(\frac{b}{a}\) then prove that a cos bθ + b sinθ = a
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 14

Question 14.
If A + B + C = 1800 than show that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
Solution:
A+B+C=180°
⇒ sin (A + B) = sin C
L.H.S = sin 2A sin 2B + sin 2C
\(=2 \sin \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \cos \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)+\sin 2 \mathrm{C}\)
= 2 sin (A+B) cos (A – b) + 2 sin C cos C
= 2 sin C cos(A-B) + 2 sinC cosC
= 2 sin C [ cos (A –  B) + cos C]
= 2 sinC [cos(A – B) – cos(A+B)
= 2 sinC (2 sin A sin B)= 4 sin A sin B sin C
= R.H.S

Question 15.
If A, B, C and angles in a triangle then prove that cos A + cos B + cos C
\(=1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 15

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 16.
If ABC are the angles in a triangle then prove that \(\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}=1+4 \sin \left(\frac{\pi-A}{4}\right) \sin \left(\frac{\pi-B}{4}\right) \sin \left(\frac{\pi-C}{4}\right)\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 16
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 17

Question 17.
Find the values of
i) sin \(\frac{5 \pi}{3}\)
ii) tan (885)°
iii) sec \(\left(\frac{13 \pi}{3}\right)\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 20

Question 18.
Simplify
(i) \(\cot \left(\theta-\frac{13 \pi}{3}\right)\)
(ii) \(\tan \left(-23 \frac{\pi}{3}\right)\)
Solution:
(i) \(\cot \left(\theta-\frac{13 \pi}{3}\right)\)
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 21

(ii) \(\tan \left(-23 \frac{\pi}{3}\right)\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 22

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 19.
Find the value of \(\begin{aligned} \sin ^2 \frac{\pi}{10} & +\sin ^2 \frac{4 \pi}{10}+ \sin ^2 \frac{6 \pi}{10}+\sin ^2 \frac{9 \pi}{10} \end{aligned}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 24

Question 20.
If sinθ= \(\frac{4}{5}\) and θ is not in the first quadrant, find the value of cos θ
Solution:
Since θ is not in the first quadrant and sin θ >0 we have 90°< θ < 180°
∴  \(\cos \theta=\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{16}{25}}=-3 / 5\)

Question 21.
If sec θ+tan θ=\(\frac{2}{3}\) find the value of sin θ and determine the quadrant in which θ lies.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 25
Since tan θ is negative, sec θ is positive
∴ θ lies in fourth quadrant.

Question 22.
Prove that \(\begin{array}{r} \cot \frac{\pi}{16} \cdot \cot \frac{2 \pi}{16} \cdot \cot \frac{3 \pi}{16} \cdots \\ \cot \frac{7 \pi}{16}=1 \end{array}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 26

Question 23.
If 3 sin θ + 4 cos θ = 5 then find the value of 4 sinθ  – 3 cos θ.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 27

Question 25.
Prove that (tanθ + cot θ)2 = sec2θ + cosec2θ = sec2 θ cosec2θ
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 29

Question 26.
If  cos θ > θ , tan θ+ sin θ = m and tan θ – sin θ = n then show that m2 – n2 = \(4 \sqrt{m n}\)
Solution:
Given that m = tan θ + sin θ
n tan θ – sin θ
∴ m+ n = 2tanθ,m – n= 2 sinθ
and (m + n)(m – n) = 4 tanθ sinθ
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 30

Question 27.
Find the rules of sin 75°, cos 75°, tan 75° and cot 75°
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 31

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 28.
Prove that \(\sin ^2 52 \frac{1}{2}^{\circ}-\sin ^2 22 \frac{1}{2}^{\circ}=\frac{\sqrt{3}+1}{4 \sqrt{2}}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 32

Question 29.
Prove that tan 700 tan 200 = 2 tan 50°
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 33

Question 30.
Express \(\sqrt{3} \sin \theta\) sinθ + cosθ as a sine of an angle
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 34

Question 31.
Prove that \(\sin ^2 \theta+\sin ^2\left(\theta+\frac{\pi}{3}\right)+\sin ^2\left(\theta-\frac{\pi}{3}\right)=\frac{3}{2}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 35

Question 32.
Let ABC be a triangle such that \(\cot A+\cot B+\cot C=\sqrt{3}\) then prove that ABC is an equilateral triangle.
Solution:
Given that \(\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ}\)
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 36
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 37

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 33.
Suppose x= tan A, y = tan B, z = tan C Suppose none of A,B,C,A-B,B-C, is an odd multiple of \(\frac{\pi}{2}\) then prove that \(\Sigma\left(\frac{x-y}{1+x y}\right)=\Pi\left(\frac{x-y}{1+x y}\right)\)
Solution:
\(\frac{1}{2}^{\circ}\) lies in first quadrant and hence all ratios are position.
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 38
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 39
Question 35.
Find the rules of
(i) \(\sin 67 \frac{1}{2}^{\circ}\)
(ii) \(\cos 67 \frac{1}{2}^{\circ}\)
(iii) \(\tan 67 \frac{1}{2}^{\circ}\)
(iv) \(\cot 67 \frac{1}{2}^{\circ}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 40

Question 36.
Simplify \(\frac{1-\cos 2 \theta}{\sin 2 \theta}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 41

Question 37.
If \(\cos A=\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}\) find the value of cos 2A
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 43

Question 38.
If \(\cos \theta=-\frac{5}{13}\) and \(\frac{\pi}{2}<\theta<\pi\) find the value of  sin 2θ
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 44

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 39.
For what values of x in the first quadrant \(\frac{2 \tan x}{1-\tan ^2 x}\) is positive?
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 45

Question 40.
If \(\cos \theta=-3 / 5\) and \(\pi<\theta<\frac{3 \pi}{2}\) find the value of \(\tan \frac{\theta}{2}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 46

Question 41.
If θ is not an integral multiple of \(\frac{\pi}{2}\) prove that tanθ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = cot θ.
Solution:
We have cotA –  tan A = 2 cot 2A
tan A = cot A – 2 cot 2A
tanθ+ 2tan2θ+4tan4θ+8cot8θ
= (cotθ  – 2 cot 2θ) +2 (cot 2θ – 2 cot 4θ) + 4 (cot 4θ – 2 cot 8θ) + 8 cot 8θ= cot θ

Question 42.
For A∈R prove that
(i) sin A sin \(\left(\frac{\pi}{3}+A\right) \sin \left(\frac{\pi}{3}-A\right)=\frac{1}{4}\) sin 3A
(ii) cos A cos \(\left(\frac{\pi}{3}+A\right) \cos \left(\frac{\pi}{3}-A\right)=\frac{1}{4}\) cos 3A
iii) sin 20° sin 40° sin 60° sin 80° = \(\frac{3}{16}\)
iv) \(\cos \frac{\pi}{9} \cos \frac{2 \pi}{9} \cos \frac{3 \pi}{9} \cos \frac{4 \pi}{9}=\frac{1}{16}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 47
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 48
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 49

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 43.
If 3A is not an odd multiple of \(\frac{\pi}{2}\), prove that tan A tan(60+A) tan(60-A) = tan 3A and hence find the value of \(\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 50

Question 44.
For α,β∈R prove that (cosα + cosβ)2 + (sinα +sinβ)2 = 4 cos2 \(\left(\frac{\alpha-\beta}{2}\right)\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 51

Question 45.
If a, b, c are non-zero real numbers and a, are solutions of the equation a cosθ + b sinθ=c then show that
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 52
Solution:
Given acosθ + bsinθ = c
a cos θ = c  b sin e
= a2 cos2θ = C2 –  2bc sin θ + b2 sin2 θ
⇒ a2 (1 – sin2 θ) = c2 – 2bc sin θ + b2 sin2 θ
⇒ (b+a) sin2 θ- 2bc sinθ + (C2 – a2) = θ
This is a quadrant equation in sin θ and suppose sin α, sin β are roots of the equation
∴ given α, β are solutions of the equation
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 53

Question 46.
If θ is not an odd multiple of \(\frac{\pi}{2}\) and \(\cos \theta=-\frac{1}{2}\) prove that \(\frac{\sin \theta+\sin 2 \theta}{1-\cos \theta+\cos 2 \theta}=\tan \theta\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 54

Question 47.
Prove that
\(\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}=\frac{3}{2}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 55

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 48.
If none of 2A and 3A is an odd multiple of \(\frac{\pi}{2}\) then prove that
tan 3A tan 2A tanA = tan 3A – tan 2A – tan A
Solution:
We have 3A = 2A+A
∴ tan 3A = tan (2A+A)
\(=\frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}\)
⇒ tan 2A + tan A tan 3A (1 – tan 2A tanA)
⇒ tan A tan2A tan 3A = tan 3A – tan 2A – tan A

Question 49.
Prove tant sin 78° + cos 132° = \(\frac{\sqrt{5}-1}{4}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 56

Question 50.
Prove tant sin 21° cos 9° – cos 84°  cos 6°  = \(\frac{1}{4}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 57

Question 51.
Find the value of sin 34° + cos 64°- cos 4°
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 58

Question 52.
Prove that cos2 76°+cos2 16°- cos 76° cos 16° \(=\frac{3}{4}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 59

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 53.
If a, b ≠ 0 and sin x+sin y=a and cos x+cos y=b find two values of
(i) \(\tan \left(\frac{x+y}{2}\right)\)
ii) \( \sin \left(\frac{x-y}{2}\right)\) is terms of a and b
Sol.
i) Given sin x+sin y=a and cos x+cos y=b we have
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 61

ii) consider
a2. b2 = (sin x + sin y)2 + (cos x + cos y)2
= sin2x + cos2 x + sin2 y cos2 y
+ 2 (sin x sin y + cos x cos y)
= 2. 2 cos (x – y)
= 2[1 .cos (x-y)]
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 62

Question 54.
Prove that cos 12°+ cos 84°+cos 132°+cos 156°
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 63

Question 55.
Show that for any 0∈ R \(4 \sin \frac{5 \theta}{2} \cos \frac{3 \theta}{2} \cos 3 \theta \) sinθ – sin 2θ+ sin 4θ +sinθ
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 64
⇒ 2 cos 3θ [ sin 4θ + sin θ]
⇒ 2 cos 3θ sin 4θ + 2 cos 3θ sin θ
⇒ sin (4θ + 3θ) sin (4θ – 3θ) + sin 4θ + sin (-2θ)
⇒ sin7θ+ sinθ + sin4θ – sin2θ
⇒ sin θ – sin 2θ+ sin 4θ – sin 7θ = R. H. S.

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 56.
If none of A, B, A + B is an integral multiple of π, then prove that
\(\begin{aligned} \frac{1-\cos A+\cos B-\cos (A+B)}{1+\cos A-\cos B-} & \cos (A+B) \\ =\tan \frac{A}{2} \cot \frac{B}{2} \end{aligned}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 65
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 66

Question 57.
For any α∈R prove that cos2
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 67
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 68

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 58.
Suppose (α-β) is not an odd multiple of \(\frac{\pi}{2}\), m is a non zero number such that m ≠ – 1 and \(\frac{\sin (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{1-m}{1+m} \quad\) then prove that \(\tan \left(\frac{\pi}{4}-\alpha\right)=m \cdot \tan \left(\frac{\pi}{4}+\beta\right)\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 69
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 70
Question 59.
If A,B, C are the angles of a triangle prove that sin 2A + sin 2B – sin 2C 4 cos A cos B sin C
Solution:
Given A, B, C are the angles of a triangle
A+B+C= π
∴ sin 2A + sin 2B – sin 2C
= sin 2A + 2cos (B + C) sin (B-C)
= sin 2A-2cosAsin (B-C)
=2 sinA cosA-2cosA sin(B – C)
= 2 cosA [sinA-sin(B  – C)]
= 2 cosA [sin(B+C)-sin(B-C)]
= 2 cos A (2 cos B sin C)
= 4 cos A cos B sin C

Question 60.
If A, B, C are angles of a triangle prove that cos 2A + cos 2B – cos 2C = 1 – 4 sin A sin B cos C
Solution:
Given A + B + C = it, we have
cos 2A + cos 2B – cos 2C
=cos2A – 2sin (B+C) sin(B-C)
cos2A – 2sinA sin(B-C)
=1 – 2 sin2A – 2sinA sin(B – C)
= 1 – 2 sinA [sin A + sin (B-C)]
= 1 – 2 sinA [sin (B+C) + sin (B-C)]
= 1 – 2 sinA (2 sin B cos C)
= 1 – 4 sin A sin B cos

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 61.
If A, B, C are angles in a triangle then prove that
(i) sin A + sin B + sin C = 4 \(\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\)
(ii) cos A + cos B + cos C = \(1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 71
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 72

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 62.
If A+B+C = \(\frac{\pi}{2}\) than show that
sin2 A+ sin2 B+ sin2 C = 1-2 sin A sin B sin C
Solution:
Given A+B + C = \(\frac{\pi}{2}\)
L.H.S = sin2 A + sin2 B + sin2 C
= sin2 A. sin2 B + 1- cos2 C
= 1 + sin2 A – (cos2 C – sin2 B)
= 1 + sin2 A-cos (C-B) cos (C-B)
=1 +sin2A-sinAcos(C-B)
= 1 +sinA [sinA – cos(B-C)]
= 1sinA [cos(B+C) – cos(B-C)]
= 1-sin A [ 2 sin B sin C]
1-2 sin A sin B sin C = R. H. S

Question 63.
If A+B+C=\(\frac{3 \pi}{2}\),prove that
cos2A+cos 2B+cos 2C = 1-4sinA sinB sinC
Solution:
LH.S. = cos2A + cos2B + cos2C
= 2cos(A+B)cos(A-B) +cos2
=-2smCcos(A-B)+ 1-2sm C
[A+B= \(\frac{3 \pi}{2}\) – C=cos(A+B)=-sinC]
=1-2 sin C[cos(A-B).sinC]
= 1-2 sin C[cos(A-B)-cos(A+B)]
= 1-2 sinC[2sinA sinB]
= 1-4 sin A sinB sinC=RH.S

Question 64.
If A,B,C are angles of a triangle then prove that
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 73
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 74
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 75

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations

Question 65.
IfA+B+C = 0 then prove that
cos2A +cos2 B + cos2C = 1+2 cosA cosB cosC
Solution:
L.H.S = cos A. cos2 B + cos2 C
= cos2 A . cos2 B. + 1 – sin.2 C
= 1 + cos2 A + cos (B + C) cos (B – C)
( ∵ A+B+C = 0 cos(B+C) = cosA)
= 1  + cos2 A + cos A cos (B – C)
= 1+ cosA [cosA+cos(B-C)]
1 – cos A [cos(B+ C) +cos(B-C)]
= 1 +cosA [2 cosB cosC]
= 1 .2 cosA cosB cosC = R.H.S

Question 66.
If A+B+C=2S then prove that
cos (S – A) + cos (S – B) + cos (S – C) + cos S = \(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations 76

TS Inter 1st Year Maths 1A Products of Vectors Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 5 Products of Vectors to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Products of Vectors Important Questions

Question 1.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}-3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) then show that \(\overline{\mathbf{a}}+\overline{\mathbf{b}}, \overline{\mathbf{a}}-\overline{\mathbf{b}}\) are perpendicular.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 1

Question 2.
If the Vectors \(\lambda \overline{\mathbf{i}}-\overline{3} \overline{\mathbf{j}}+5 \overline{\mathrm{k}}, 2 \lambda \overline{\mathrm{i}}-\lambda \overline{\mathbf{j}}-\overline{\mathrm{k}}\) are perpendicular to each other find λ.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 2

Question 3.
If \(\overline{\mathbf{a}}\) =6 \(\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+3 \overline{\mathrm{k}}\) and \(\overline{\mathbf{b}}=2 \overline{\mathbf{i}}-9 \overline{\mathrm{j}}+6 \overline{\mathrm{k}}\) then find \(\overline{\mathrm{a}}, \overline{\mathrm{b}}\) and the angle between \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 3
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 4

Question 4.
Let \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+4 \overline{\mathbf{j}}-5 \overline{\mathrm{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathrm{k}}\) and \( \overline{\mathbf{c}}=\overline{\mathbf{i}}+2 \overline{\mathrm{k}}\). Find unit vector in the opposite direction of \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 5

Question 5.
Let \(\bar{a}\) and \(\bar{b}\) be non zero, non collinear vectors. If \(|\overline{\mathbf{a}}+\bar{b}|=|\overline{\mathbf{a}}-\bar{b}|\) then find the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 6
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 7

Question 6.
If \(|\overline{\mathbf{a}}|=11,|\bar{b}|=23 \text { and }|\overline{\mathbf{a}}-\overline{\mathbf{b}}|\) = 30 then find the angle between the vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\), and also find \(|\overline{\mathbf{a}}+\overline{\mathbf{b}}|\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 8

Question 7.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-\overline{\mathbf{j}}-\overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}-3 \overline{\mathbf{j}}+\overline{\mathbf{k}}\) then find the projection vector of \(\overline{\mathbf{b}} \text { on } \bar{a}\) and its magnitude.
Solution:
Projection vector of \(\overline{\mathbf{b}} \text { on } \bar{a}\) is
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 9

Question  8.
If P, Q, R, S are points whose position vectors are \(\overline{\mathbf{i}}-\overline{\mathbf{k}},-\overline{\mathbf{i}}+\mathbf{2} \overline{\mathbf{j}}, 2 \overline{\mathbf{i}}-3 \overline{\mathbf{k}}\) and 3 \(\overline{\mathbf{i}}-2 \overline{\mathbf{j}}-\overline{\mathbf{k}}\) respectively, then find the component of \(\overline{\mathbf{R S}}\) on \(\overline{\mathbf{P Q}}\).
Solution:
Let O be the origin.
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 10

Question 9.
Let \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+3 \overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=4 \overline{\mathbf{i}}+\overline{\mathbf{j}}\) and \(\bar{c}=\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-7 \overline{\mathbf{k}}\). Find the vector \(\overline{\mathbf{r}}\) such that \(\overline{\mathbf{r}} \cdot \overline{\mathbf{a}}=9, \overline{\mathbf{r}} \cdot \overline{\mathbf{b}}=7 \text { and } \overline{\mathbf{r}} \cdot \overline{\mathbf{c}}=6\)
Solution:
Let \(\bar{r}=x \bar{i}+y \bar{j}+z \bar{k}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 11
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 12

Question 10.
Show that the points \(2 \bar{i}-\bar{j}+\bar{k}, \bar{i}-3 \bar{j}-5 \bar{k}\) and \( 3 \bar{i}-4 \bar{j}-4 \bar{k}\) are the vertices of a right angled triangle. Also find the other angles.
Solution:
Let O be the origin and A,B,C  be the given points, then
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 13
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 14

Question 11.
Prove that the smaller angle O between any two diagonals of a cube is given by \(\cos ^{-1}\left(\frac{1}{3}\right)\)
Solution:
Consider a unit cube with its vertices as shown in the figure
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 15
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 16
Similarly we can show the result for any other two diagonals of the cube.

Question 12.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) be non zero mutually orthogonal
vectors if \(\mathbf{x} \overline{\mathbf{a}}+\mathbf{y} \overline{\mathbf{b}}+\mathbf{z} \overline{\mathbf{c}}=\overline{\mathbf{0}}\) then x = y = z = 0
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 17

Question 13.
If \(4 \bar{i}+\frac{2 p}{3} \bar{j}+p \bar{k}\) is parallel to the vector \(\overline{\mathbf{i}}+2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}\). find p.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 18
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 19

Question 14.
Let \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}} \) be vectors satisfying \(|\overline{\mathbf{a}}|=|\bar{b}|=5\) and \((\overline{\mathbf{a}}, \overline{\mathbf{b}})=45^{\circ}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 20

Question 15.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) be mutually orthogonal vectors of equal magnitudes. Prove that the vector \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\) is equally inclined to each of \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\), the angle of inclination being \(\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 21
Similarly we can show the result for any other two diagonals of the cube.

Question 16.
The vectors \(\overline{\mathrm{AB}}=3 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) and \(\overline{\mathrm{AD}}=\overline{\mathrm{i}}-2 \overline{\mathrm{k}}\) represent the adjacent sides of a parallelogram A B C D. Find the angle between the diagonals.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 22
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 23

Question 17.
For any two vectors \(\bar{a}\) and \(\bar{b}\) show that
(i) \(|\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}| \leq|\overline{\mathbf{a}}||\overline{\mathbf{b}}|\) (Cauchy – Schawartz in equality)
(ii) \(|\overline{\mathbf{a}}+\overline{\mathbf{b}}| \leq|\overline{\mathbf{a}}|+|\overline{\mathbf{b}}|\)(Triangle inequality)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 25

Question 18.
Find the area of parallelogram for which the vectors \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-3 \overline{\mathbf{j}} \text { and } \overline{\mathbf{b}}=3 \overline{\mathbf{i}}-\overline{\mathbf{k}}\) are adjacent sides.
Solution:
The vector area of the parallelogram
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 26
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 27

Question 19.
Let \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=\mathbf{2} \overline{\mathbf{i}}-\overline{\mathbf{j}}+\mathbf{3} \overline{\mathbf{k}}, \overline{\mathbf{c}}=\overline{\mathbf{i}}-\overline{\mathbf{j}}\) and \(\bar{d}=6 \bar{i}+2 \bar{j}+3 \bar{k}\). Express \(\bar{d}\) interms of \(\overline{\mathbf{b}} \times \overline{\mathbf{c}}, \overline{\mathbf{c}} \times \overline{\mathbf{a}}\) and \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 28
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 29
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 30
Question 20.
Show that the angle in a semicircle is a right angle.
Solution:
Let O be the centre and AOB be the diameter of the given semicircle. Let P be any point on it. Let the position vectors of A and P be taken as \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{p}}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 31
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 32

Question 21.
For any vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) prove that \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}=(\overline{\mathbf{a}} \cdot \overline{\mathbf{c}}) \overline{\mathbf{b}}-(\overline{\mathbf{b}} \cdot \overline{\mathbf{c}}) \overline{\mathbf{a}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 33
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 34

Question 22.
Find the cartesian equation of the plane passing through the point (-2,1,3) and perpendicular to the vector 3 \(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\mathbf{5} \overline{\mathbf{k}}\).
Solution :
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 35

Question 23.
Find the cartesian equation of the plane through the point A (2, – 1, -4) and parallel to the plane 4x – 12y – 3z – 7 = 0.
Solution:
The equation of the parallel plane is 4x- 12y-3z = p
11 passes through A (2, – 1, – 4) then
4(2)-12 (- 1)-3(-4) = p
⇒ 8 + 12 + 12 = p = p = 32
The equation of the required plane is
4x- 12y-3z = 32

Question 24.
Find the angle between the planes
2x – 3y – 6z = 5 and 6x + 2y – 9z = 4.
Solution:
Equation of the plane is 2x – 3y – 6z = 5
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 36

Question 25.
Find limit vector orthogonal to the vector \(3 \bar{i}+2 \bar{j}+6 \bar{k}\) and coplanar with the vectors
\(2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}} \text { and } \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 37
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 38

Question 26.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-3 \overline{\mathbf{j}}+5 \overline{\mathbf{k}}, \overline{\mathbf{b}}=-\overline{\mathbf{i}}+4 \overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) then find \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}\) and unit vector perpendicular to both \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 39

Question 27.
If \(\overline{\mathbf{a}}=2 \overline{\mathrm{i}}-3 \overline{\mathrm{j}}+5 \overline{\mathrm{k}}, \overline{\mathrm{b}}=-\overline{\mathrm{i}}+4 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) then find \((\bar{a}+\bar{b})_{\times}(\bar{a}-\bar{b})\) and unit vector perpendicular to both \(\overline{\mathbf{a}}+\overline{\mathbf{b}}\) and \(\overline{\mathbf{a}}-\overline{\mathbf{b}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 40

Question 28.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}, \overline{\mathbf{d}}\) are vectors such that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{c}} \times \overline{\mathbf{d}}\) and \(\overline{\mathbf{a}} \times \overline{\mathbf{c}}=\overline{\mathbf{b}} \times \overline{\mathbf{d}}\) then show that the vectors \(\overline{\mathbf{a}}-\overline{\mathbf{d}}\) and \(\overline{\mathbf{b}}-\overline{\mathbf{c}}\) are parallel.
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 41

Question 29.
Find the area of the triangle formed by the two sides \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=3 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}-\overline{\mathbf{k}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 42

Question 30.
In a \(\triangle \mathrm{ABC}\) if \(\overline{\mathrm{BC}}=\overline{\mathrm{a}}, \overline{\mathrm{CA}}=\overline{\mathrm{b}}\) and \(\overline{\mathrm{AB}}=\overline{\mathbf{c}}\) then show that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{b}} \times \overline{\mathbf{c}}=\overline{\mathbf{c}} \times \overline{\mathbf{a}} \cdot\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 43

Question 31.
Let \(\overline{\mathbf{a}}=\mathbf{2} \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=3 \overline{\mathbf{i}}+4 \overline{\mathbf{j}}-\overline{\mathbf{k}}\). If ‘θ’ is the angle between \(\bar{a}\) and \(\bar{b}\) then find \(\sin \theta\)
Solution:
\(\bar{a}=2 \bar{i}-\bar{j}+\bar{k}, \bar{b}=3 \bar{i}+4 \bar{j}-\bar{k}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 44

Question 32.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) be such that \(\overline{\mathbf{c}} \neq \overline{\mathbf{0}}, \overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{c}}, \overline{\mathbf{b}} \times \overline{\mathbf{c}}=\overline{\mathbf{a}}\). Show that \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are pair wise orthogonal vectors and \(|\overline{\mathbf{b}}|=1,|\overline{\mathbf{c}}|=|\overline{\mathbf{a}}|\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 49
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 45

Question 33.
Let \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-2 \overline{\mathbf{k}}\); \(\overline{\mathbf{b}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}\). If \(\bar{c}\) is a vector such that \(\overline{\mathbf{a}} \cdot \overline{\mathbf{c}}=|\overline{\mathbf{c}}|,|\overline{\mathbf{c}}-\overline{\mathbf{a}}|=2 \sqrt{2}\) and the angle between \(\overline{\mathbf{a}} \times \overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) is 30° then find the value of \(|(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times \overline{\mathbf{c}}|\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 46

Question 34.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) be two non-collinear unit vectors if \(\bar{\alpha}=\overline{\mathbf{a}}-(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}) \overline{\mathbf{b}}\) and \(\bar{\beta}=\overline{\mathbf{a}} \times \overline{\mathbf{b}}\) then show that \(|\bar{\beta}|=|\bar{\alpha}|\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 50

Question 35.
A non zero vector \(\overline{\mathbf{a}}\) is parallel to the line of intersection of the plane determined by the vectors \(\overline{\mathbf{i}}, \overline{\mathbf{i}}+\overline{\mathbf{j}}\) and the plane determined by the vectors \(\overline{\mathbf{i}}-\overline{\mathbf{j}}, \overline{\mathbf{i}}+\overline{\mathbf{k}}\). Find the angle between \(\bar{a}\) and the vector \(\bar{i}-2 \bar{j}+2 \bar{k}\).
Solution:
Let l be the line of intersection of planes determined by the pairs \(\bar{i}, \bar{i}+\bar{j}\) and
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 51
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 52

Question 36.
Let \(\overline{\mathbf{a}}=4 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}-\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}=\overline{\mathbf{i}}-4 \overline{\mathbf{j}}+5 \overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=3 \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\). Find the vector α which is perpendicular to both \(\bar{a}\) and \(\bar{b}\) and \(\alpha \cdot \overline{\mathbf{c}}\) = 21 \(\bar{\alpha}\) is perpendicular to both \(\bar{a}\) and \(\bar{b}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 53

Question 37.
For any vector \(\overline{\mathbf{a}}\) show that
\(|\overline{\mathbf{a}} \times \overline{\mathbf{i}}|^2+|\overline{\mathbf{a}} \times \overline{\mathbf{j}}|^2+|\overline{\mathbf{a}} \times \overline{\mathbf{k}}|^2=2|\overline{\mathbf{a}}|^2\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 54

Question 38.
If \(\bar{a}\) is a non zero vector and \(\bar{b}\) and \(\bar{c}\) are two vectors such that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{a}} \times \overline{\mathbf{c}}\) and \(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\overline{\mathbf{a}} \cdot \overline{\mathbf{c}}\) then prove that \(\overline{\mathbf{b}}=\overline{\mathbf{c}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 55

Question 39.
Prove that the vectors \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=3 \overline{\mathbf{i}}-4 \overline{\mathbf{j}}-4 \overline{\mathbf{k}}\) are coplanar.
Solution:
\(\bar{a}=2 \bar{i}-\bar{j}+\bar{k}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 56

Question 40.
Find the volume of the parallelopiped whose coterminous edges are represented by the vectors \(2 \bar{i}-3 \bar{j}+\bar{k}, \bar{i}-\bar{j}+2 \bar{k}\) and \(\mathbf{2} \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 57

Question 41.
Show that
\(\overline{\mathbf{i}} \times(\overline{\mathbf{a}} \times \overline{\mathbf{i}})+\overline{\mathbf{j}} \times(\overline{\mathbf{a}} \times \overline{\mathbf{j}})+\overline{\mathbf{k}} \times(\overline{\mathbf{a}} \times \overline{\mathbf{k}})=2 \overline{\mathbf{a}}\) For any vector \(\overline{\mathbf{a}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 58

Question 42.
Prove that for any three vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}} \left[\begin{array}{lll}
\bar{b}+\bar{c} & \bar{c}+\overline{\mathbf{a}} & \overline{\mathbf{a}}+\bar{b}
\end{array}\right]=2\left[\begin{array}{lll}
\overline{\mathbf{a}} & \bar{b} & \overline{\mathbf{c}}
\end{array}\right]\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 59

Question 43.
For any three vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) prove that
\(\left[\begin{array}{lll}
\overline{\mathbf{b}} \times \overline{\mathbf{c}} & \overline{\mathbf{c}} \times \overline{\mathbf{a}} & \overline{\mathbf{a}} \times \overline{\mathbf{b}}
\end{array}\right]=\left[\begin{array}{lll}
\overline{\mathbf{a}} & \overline{\mathbf{b}} & \overline{\mathbf{c}}
\end{array}\right]^2 \cdot(\mathbf)\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 60

Question 44.
Let \(\bar{a}, \bar{b}\) and \(\bar{c}\) be unit vectors such that \(\bar{b}\) is not parallel to \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}})=\frac{1}{2} \overline{\mathbf{b}}\). Find the angles made by \(\bar{a}\) with each of \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 61

Question 45.
For any four vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\).
Prove that \((\bar{b} \times \overline{\mathbf{c}}) \cdot(\overline{\mathbf{a}} \times \overline{\mathbf{d}})+(\overline{\mathbf{c}} \times \overline{\mathbf{a}}) \cdot(\overline{\mathbf{b}} \times \overline{\mathrm{d}}) +(\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \cdot(\overline{\mathbf{c}} \times \overline{\mathbf{d}})=0(\mathrm)\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 62
= 0

Question 46.
Find the equation of the plane passing through the points \(\mathrm{A}=(2,3,-1), \mathrm{B}=(4,5, 2)\) and C=(3,6,5).
Solution:
Let \(\overline{\mathrm{OA}}=2 \overline{\mathrm{i}}+3 \overline{\mathrm{j}}-\overline{\mathrm{k}}
\overline{\mathrm{OB}}=4 \overline{\mathrm{i}}+5 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\)
\(\overline{\mathrm{OC}}=3 \overline{\mathrm{i}}+6 \overline{\mathrm{j}}+5 \overline{\mathrm{k}}\) with respect to origin \(\mathrm{O}\).
Let P be any point on the plane passing through the points A,B,C
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 63

Question 47.
Find the equation of the plane passing through the point A(3,-2,-1) and parallel to the vectors \(\bar{b}=\bar{i}-2 \bar{j}+4 \bar{k}\) and \(\overline{\mathbf{c}}=3 \overline{\mathbf{i}}+2 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\).
Solution:
The equation of the plane passing through A=(3,-2,-1) and parallel to the vectors
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 64
Question 48.
Find the vector equation of the plane passing through the intersection of planes \(\overline{\mathbf{r}} \cdot(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}})=6\) and \(\overline{\mathbf{r}} \cdot(2 \bar{i}+3 \overline{\mathbf{j}}+4 \overline{\mathbf{k}})=-5\) and the point (1,1,1).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 66
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 65

Question 49.
Find the distance of the point (2,5,-3) from the plane \(\overline{\mathbf{r}} \cdot(6 \bar{i}-3 \bar{j}+2 \bar{k})=4 \cdot\)
Solution:
Here \(\bar{a}=\bar{i}+5 \bar{j}-3 \bar{k}, \bar{n}=6 \bar{i}-3 \bar{j}+2 \bar{k}\) and d=4
Then \(\overline{\mathrm{r}} \cdot \overline{\mathbf{n}}=\overline{\mathrm{d}}\)
The distance of the point (2,5,-3) from the given plane is
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 67

Question 50.
Find the angle between the line \(\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}\) and the plane 10 x+2 y-11 z=3
Solution:
Let φ be the angle between the given line and normal to the plane.
Concert the above equations to vector notation,
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 68

Question 51.
For any four vectors \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\) show that
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 69
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 70

Question 52.
Find the shortest distance between the skew
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 71
Solution:

TS Inter 1st Year Maths 1A Products of Vectors Important Questions 80

The first line passes through the point A(6,2,2) and parallel to the vector \(\overline{\mathrm{b}}=\overline{\mathrm{i}}-2 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\).
The second line passes through the point C(-4,0,-1) and parallel to the vector \(\overline{\mathrm{d}}=3 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}-2 \overline{\mathrm{k}}\)
Shortest distance is =\(\frac{|[\overline{\overline{A C}} \bar{b} \bar{d}]|}{|\bar{b} \times \bar{d}|}\)

TS Inter 1st Year Maths 1A Products of Vectors Important Questions 72

Question 53.
i) Show that the altitudes of a triangle are concurrent.
ii) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:
(i)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 73
Consider ΔABC . O is point of intersection of altitudes.
To prove that the three altitudes are concurrent at ‘ O ‘. We have to prove that \(\overline{\mathrm{OC}}\) is perpendicular to \(\overline{\mathrm{AB}}\)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 81
∴  \(\overline{\mathrm{OC}}\) is the third altitude which passes through ‘ O ‘.
Hence the three altitudes of the triangle are concurrent.

(ii)
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 82
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 75
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 76

Question 54.
Show that the vector area of the quadrilateral ABCD having diagonals \(\overline{\mathrm{AC}}, \overline{\mathrm{BD}}\) is \(\frac{1}{2}(\overline{\mathrm{AC}} \times \overline{\mathrm{BD}})\)
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 77
ABCD is a quadrilateral. \(\overline{\mathrm{AC}}\) and \(\overline{\mathrm{BD}}\) are diagonals of the quadrilateral. Q is the point of intersection of diagonals.
Vector area of quadrilateral ABCD = Sum of the vector area of ΔAQB, ΔBQC, ΔCQD and ΔDQA.TS Inter 1st Year Maths 1A Products of Vectors Important Questions 83

Question 55.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) be unit vectors such that \(\bar{b}\) is not parallel to \(\bar{c}\) and \(\bar{a} \times(\bar{b} \times \bar{c})=\frac{1}{2} \bar{b}\). Find the angle made by the vector \(\bar{a}\) with each of the vectors \(\bar{b}\) and \(\bar{c}\).
Solution:
TS Inter 1st Year Maths 1A Products of Vectors Important Questions 79

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 4 Addition of Vectors to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions

Question 1.
Show that the points whose position vectors are \(-2 \overline{\mathbf{a}}+3 \overline{\mathbf{b}}+5 \overline{\mathbf{c}}, \overline{\mathbf{a}}+2 \overline{\mathbf{b}}+3 \overline{\mathbf{c}}, 7 \overline{\mathbf{a}}-\overline{\mathbf{c}}\) are collinear when, \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non coplanar vectors.
Solution:
Let O be the origin of reference so that
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 1

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions

Question 2.
If A B C D E F is a regular hexagon with centre o then prove that \(\overline{\mathrm{AB}}+\overline{\mathrm{AC}}+\overline{\mathrm{AD}}+\overline{\mathrm{AE}}+\overline{\mathrm{AF}}= \overline{3 \mathrm{AD}}=6 \overline{\mathrm{AO}}\).
Solution:
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 3

Question 3.
In the two dimensional plane, prove by using vector methods, the equation of the line whose intercepts on the axes are a and b is \(\frac{x}{a}+\frac{y}{b}\) = 1
Solution:
Let A = (a,0), B = (0,b). P = (x,y)
Let O be the origin so that
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 4

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions

Question 4.
Find a unit vector in the direction of the vector = \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+3 \overline{\mathbf{j}}+\overline{\mathrm{k}}\)
Solution:
The unit vector in the direction of the vector
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 5

Question 5.
Find a vector in the direction of vector \(\overline{\mathbf{a}}=\overline{\mathbf{i}}-2 \overline{\mathbf{j}}\) that has magnitude 7 units.
Solution:
The unit vector in the direction of given
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 6

Question 6.
Find the unit vector in the direction of sum of the vectors \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+2 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\) and \( \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+3 \overline{\mathbf{k}}\)
Solution:
The sum of the vectors is
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 7

Question 7.
Write the direction ratios of the vector \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}-\mathbf{2} \overline{\mathbf{k}}\) and hence calculate its direction cosines.
Solution:
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 8

Question 8.
Consider the two points P and Q with position vectors \(\overline{\mathrm{OP}}=3 \overline{\mathrm{a}}-2 \overline{\mathrm{b}}\) and \(\overline{\mathrm{OQ}}= \overline{\mathbf{a}}+\overline{\mathbf{b}}\). Find the position vector of a point R which divides the line joining P and Q in the ratio 2: 1 (i) internally and (ii) externally.
Solution:
i) The position vector of the point R dividing the joining of  P and Q internally in the ratio 2: 1 is
\(\overline{\mathrm{OR}}=\frac{2(\overline{\mathrm{a}}+\overline{\mathrm{b}})+(3 \overline{\mathrm{a}}-2 \overline{\mathrm{b}})}{2+1}=\frac{5 \bar{a}}{3}\)

ii) The position vector of the point R dividing the joining of P and Q externally in the ratio 2: 1 is
\(\overline{\mathrm{OR}}=\frac{2(\overline{\mathrm{a}}+\overline{\mathrm{b}})-(3 \overline{\mathrm{a}}-2 \overline{\mathrm{b}})}{2-1}=4 \overline{\mathrm{b}}-\overline{\mathrm{a}}\)

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions

Question 9.
Show that the points \(A(2 \bar{i}-\bar{j}+\bar{k}), \dot{B}(\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}), \mathbf{C}(3 \overline{\mathbf{i}}-4 \overline{\mathbf{j}}-4 \overline{\mathbf{k}})\) are the vertices of a right angled triangle.
Solution:
We have \(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\)
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 9
∴ AB2 = BC2 + CA2 and hence a right angled triangle can be formed with the points A, B, and C.

Question 10.
Show that the points \(\mathbf{A}(2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}) \dot{B}(\bar{i}-3 \bar{j}-5 \bar{k}), C(3 \bar{i}-4 \bar{j}-4 \bar{k})\) are the vertices of a right angled triangle.
Solution:
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 11

Question 11.
In a ΔABC if \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are the position vectors of the vertices A, B and C respectively, then prove that the position vector of the centroid G is \(\frac{1}{3}(\bar{a}+\bar{b}+\bar{c})\).
Solution:

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 13

Let G be the centroid of ΔABC and AD is the median through the vertex A.
Then AG : GD = 2: 1
Suppose \(\overline{\mathrm{OA}}=\overline{\mathrm{a}}, \overline{\mathrm{OB}}=\overline{\mathrm{b}}, \overline{\mathrm{OC}}=\overline{\mathrm{c}}\) with
reference to the specific origin O.
Mid point of BC is = \(\overline{\mathrm{OD}}=\frac{1}{2}(\overline{\mathrm{b}}+\overline{\mathrm{c}})\)
Since G divides AD in the ratio 2: 1 we have
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 14

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions

Question 12.
In a ΔABC, If ‘O’ Is the circumcentre, and H is the orthocentre then show that
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 15
Solution:
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 16
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 17

Question 13.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}, \overline{\mathbf{d}}\) be the position vectors of A, B, C and D respectively which are the vertices of a tetrahedron. Then prove that the lines joining the vertices to the centroids of the opposite faces are concurrent. (this point is called the centroid or the centre of the tetrahedron)
Solution:
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 18

Let O be the origin and G1, G2, G3, G4 be the centroids of ΔBCD, ΔCAD, ΔABD and ΔABC.
Then \(\overline{\mathrm{OG}}_1=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}+\overline{\mathrm{d}}}{3}\)
Suppose P is the point which divides AG1 in the ratio 3: 1 then
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 19
Similarly position vectors of the points dividing BC2, CG3 and DG4 in the ratio 3: 1 are each equal to \(\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}+\overline{\mathrm{d}}}{4}\)
Hence the point P lies on AG1, BC2, CG3, DG4.

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions

Question 14.
Let OABC be a parallelogram and D is the midpoint of \(\overline{\mathbf{O A}}\). Prove that the segment is \(\overline{\mathbf{C D}}\) trisects the diagonal \(\overline{\mathbf{O B}}\) and is trisected by the diagonal\(\overline{\mathbf{O B}}\)
Solution:
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 20
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 21

Question 15.
Let \(\overline{\mathbf{a}}, \overline{\mathbf{b}}\) be non-collinear vectors, if
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are such that 3α = 2β then find x and y.
Solution:
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 23

Question 16.
If the points whose position vectors are \(3 \overline{\mathbf{i}}-2 \overline{\mathbf{j}}-\overline{\mathbf{k}}, \quad 2 \overline{\mathbf{i}}+3 \overline{\mathbf{j}}-4 \overline{\mathbf{k}},-\overline{\mathbf{i}}+\overline{\mathbf{j}}+2 \mathbf{k} 4 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}+\lambda \overline{\mathbf{k}}\) are coplanar, then show that \(\lambda=\frac{-146}{17}\)
Solution:
Let O be the origin and let A, B, C and D be the given points. Then
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 25
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 26

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions

Question 17.
Find the equation of the line parallel to the vector \(2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) and which passes through the point A whose position vector is \(2 \bar{i}-\bar{j}+2 \bar{k}\). If P is a point on this line such that AP = 15, find the position vector of P.
Solution:
The vector equation of the line passing through the point \(\mathrm{A}(\bar{a})\) whose position vector is \(\overline{\mathrm{a}}=3 \overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}\) and parallel to the vector \(\overline{\mathrm{b}}=2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\mathrm{t} \overline{\mathrm{b}}\) for some t ∈ R
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 27

Question 18.
Show that the line joining the pair of points \(6 \bar{a}-4 \bar{b}+4 \bar{c},-4 \bar{c}\) and the line joining the pair of points \(-\bar{a}-2 \bar{b}-3 \bar{c}, \bar{a}+2 \bar{b}-5 \bar{c}\) intersect at the point \(-4 \bar{c}\) when \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non coplanar vectors.
Solution:
The vector equation of the line joining points
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 30
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 28
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 29
and equation (5) is satisfied.
∴ The two lines intersect at the point from (1) is \(-4 \bar{c}\)

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions

Question 19.
Find the point of intersection of the line \(\overline{\mathbf{r}}=\mathbf{2} \overline{\mathbf{a}}+\overline{\mathbf{b}}+\mathbf{t}(\overline{\mathbf{b}}-\overline{\mathbf{c}})\) and the plane \(\overline{\mathbf{r}}=\overline{\mathbf{a}}+\mathbf{x}(\overline{\mathbf{b}}+\overline{\mathbf{c}})+\mathbf{y}(\overline{\mathbf{a}}+2 \overline{\mathbf{b}}-\overline{\mathbf{c}})\) where a, b, c are non-coplanar vectors.
Solution:
Let \(\overline{\mathbf{r}}\) be the position vector of the point P the intersection of the line and the plane.
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Question 20.
Prove that the vector equation of line through the points \(A(\bar{a}), B(\bar{b})\) is \(\overline{\mathbf{r}}=(\mathbf{1}-\mathbf{t}) \overline{\mathbf{a}}+\mathbf{t} \overline{\mathbf{b}}, \mathbf{t} \in \mathbf{R}\).
Solution:
Let O be the origin and P be any point on the line.
TS Inter 1st Year Maths 1A Addition of Vectors Important Questions 32