Students must practice these TS Intermediate Maths 1B Solutions Chapter 3 Straight Lines Ex 3(b) to find a better approach to solving the problems.
TS Inter 1st Year Maths 1B Straight Lines Solutions Exercise 3(b)
I.
Question 1.
Find the sum of squares of the intercepts of the line 4x – 3y = 12 on the axes of coordinates. (VS.A.Q.)
Answer:
Equation of the given line is 4x – 3y = 12
Writing in the intercepts form \(\frac{x}{a}+\frac{y}{b}\) = 1
∴ a = 3, b = – 4
∴ Sum of the squares = a2 + b2 = 9 + 16 – 25
Question 2.
If the portion of a straight line intercepted between the axes of coordinates is bisected at (2p, 2q), write the equation of the straight (V.S.A.Q.)
Answer:
Coordinates of A = (a, 0) and B = (0, b)
M is the midpoint of AB
∴ M = \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) = (2p, 2q)
∴ \(\frac{a}{2}\) = 2p and \(\frac{b}{2}\) = 2q ⇒ a = 4p, b = 4q
Question 3.
If the linear equations ax + by + c = 0 (abc ≠ 0) and lx + my + n = 0 represents the same line and r = \(\frac{l}{a}\) = \(\frac{n}{c}\), write the value of r in terms of m and b. (V.S.A.Q.)
Answer:
ax + by + c = 0 and lx + my + n = 0 represents the same line.
Hence coefficients are proportional
∴ \(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}\) = r
r = \(\frac{\mathrm{m}}{\mathrm{b}}\) is the value of ’r’ interms of m and b.
Question 4.
Find the angle made by the straight line y = – √3 x + 3 with the positive direction of the X – axis measured in the counter clock-wise direction. (V.S.A.Q.)
Answer:
Equation of the given line is y = – √3x + 3 which is of the form y = mx + c where
m = tan α = √3 = tan \(\frac{2 \pi}{3}\) (angle made by the line in counter clock-wise direction)
∴ α = \(\frac{2 \pi}{3}\)
Question 5.
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. (V.S.A.Q.)
Answer:
Equation of the line in the intercepts form is
\(\frac{x}{a}+\frac{y}{b}\) – 1 = 0
p = length of the perpendicular drawn from origin to the line then
II.
Question 1.
In what follows, p denotes the distance of the straight line from the origin and a denotes the angle made by the normal ray drawn from the origin to the straight line with \(\overleftrightarrow{O x}\) measured in the anti-clockwise sense. Find the equations of the straight lines with the following values of p and α. (V.S.A.Q.)
(i) p = 5, α = 60°
(ii) p = 6, α = 150°
(iii) p = 1, α = \(\frac{7 \pi}{4}\)
(iv) p = 4, α = 90°
(v) p = 0, α = 0
vi) P = 2√2, α = \(\frac{5 \pi}{4}\)
Answer:
Equation of line in the normal form is x cos α + y sin α = p
(i) p = 5, α = 60°
Equation of line is x cos 60°+ y sin 60°= 5
⇒ \(\frac{x}{2}\) + y\(\frac{\sqrt{3}}{2}\) = 5
⇒ x + √3y = 10
(ii) p = 6, α = 150°
Equation of line is x cos 150° + y sin 150°= 6
⇒ x cos ( 180° – 30°) + y sin (180° -30°) = 6
⇒ – x cos 30° + y sin 30° = 6
⇒ – x \(\frac{\sqrt{3}}{2}\) + y \(\frac{1}{2}\) = 6
⇒ √3x + y = 12
⇒ √3x – y + 12 = 0
(iii) P = 1, α = \(\frac{7 \pi}{4}\)
Equation of the line is x cos \(\frac{7 \pi}{4}\) + y sin \(\frac{7 \pi}{4}\) = 1
⇒ x cos 315° + y sin 315° = 1
⇒ x cos (360° – 45°) + y sin (360° – 45°) = 1
⇒ x cos 45° – y sin 45° = 1
⇒ \(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\) = 1
⇒ x – y – √2 = 0
(iv) p = 4, α = 90°
Equation of the line is
x cos 90° + y sin 90° = 4
⇒ x(0) + y(1) = 4
⇒ y – 4 = 0
(v) p = 0, α = 0
Equation of the line is
x cos 0 + y sin 0 = 0
⇒ x = 0
(vi) p = 2√2 , α = \(\frac{5 \pi}{4}\)
Equation of the line is
x cos \(\frac{5 \pi}{4}\) + y sin \(\frac{5 \pi}{4}\) = 2√2
⇒ x cos 225° + y sin 225° = 2√2
⇒ x cos (180° + 45°) + y sin( 180° + 45°) = 2√2
⇒ – x cos 45° – y sin 45° = 2√2
⇒ – \(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\) = 2√2
⇒ x + y + 4 = 0
Question 2.
Find the equations of the straight lines in the symmetric form, given the slope and a point on the line in each part of the question. (S.A.Q.)
(i) √3, (2, 3)
(ii) – \(\frac{1}{\sqrt{3}}\), (- 2, 0)
(iii) – 1, (1, 1)
Answer:
(i) Equation of the line in symmetric form is
\(\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}\) = r
(x1, y1) = (2, 3), slope m = tan θ = √3 ⇒ θ = 60°
∴ Equation of the line is
\(\frac{x-2}{\cos \theta}=\frac{y-3}{\sin \theta}\) = √3
⇒ \(\frac{x-2}{\cos 60^{\circ}}=\frac{y-3}{\sin 60^{\circ}}\)
(ii) (x1, y1) = (-2, 0)
Answer:
Slope m = tan θ = – \(\frac{1}{\sqrt{3}}\)
⇒ θ = 180° – 30° = 150°
∴ Equation of the line is \(\frac{x+2}{\cos 150^{\circ}}=\frac{y}{\sin 150^{\circ}}\)
(iii) tan α = – 1, α = 180 – 45°
(x1, y1) = (1, 1)
∴ Equation of the line is
\(\frac{x-1}{\cos \left(\frac{3 \pi}{4}\right)}\) = \(\frac{y-1}{\sin \left(\frac{3 \pi}{4}\right)}\)
Question 3.
Transform the following equations into
(a) slope – intercept form (b) intercept form and (c) normal form. (S.A.Q.)
(i) 3x + 4y = 5
(ii) 4x – 3y + 12 = 0,
(iii) √3x + y = 4
(iv) x + y + 2 = 0
(v) x + y – 2 = 0
(vi) √3x + y + 10 = 0
Answer:
(i) 3x + 4y = 5
(a) Slope – intercept form : 4y = 5 – 3x
⇒ y = \(\frac{5}{4}-\frac{3}{4}\)
Slope m = – \(\frac{3}{4}\) and Y- intercept = \(\frac{5}{4}\)
(b) Intercept form : 3x + 4y = 5
⇒ \(\frac{x}{\left(\frac{5}{3}\right)}+\frac{y}{\left(\frac{5}{4}\right)}\) = 1
This is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1
(c) Normal form: 3x + 4y = 5,
Dividing by \(\sqrt{9+16}\) = 5 on both sides
(ii) 4x – 3y + 12 = 0,
(a) Slope – intercept form: 3y = 4x + 12
⇒ y = \(\left(\frac{4}{3}\right)\)x + 4
Slope = \(\frac{4}{3}\) and y-intercept = 4
(b) Intercept form: 4x – 3y = – 12
⇒ \(\frac{4}{-12} x+\frac{3}{12} y\) = 1
⇒ \(\frac{x}{(-3)}+\frac{y}{(4)}\) = 1
(c) Normal form : 4x – 3y + 12 = 0
⇒ – 4x + 3y = 12
Dividing by \(\sqrt{16+9}\) = 5 on both sides
(iii) √3x + y = 4
Answer:
(a) Slope – intercept form : y = -√3x + 4
Slope = – √3 , Y-intercept = 4
(b) Intercept form: √3 x + y = 4
(c) Normal form: √3x + y = 4
Dividing by √3 + 1 = 2 on both sides
(iv) x + y + 2 = 0
(a) Slope – intercept form: y = – x – 2
Slope = – 1, Y – intercept = – 2,
(b) Intercept form: x + y = – 2
⇒ \(\frac{x}{-2}+\frac{y}{-2}\) = 1
(c) Normal form : x + y + 2 = 0
⇒ – x – y = 2
Dividing by \(\sqrt{1+1}\) = √2 on both sides
(v) x + y – 2 = 0
(a) Slope – intercept form: y = – x + 2
Slope = -1, Y – intercept = 2
(b) Intercept form: x + y -2 = 0
⇒ x + y = 2
⇒ \(\frac{x}{2}+\frac{y}{2}\) = 1
(c) Normal form: x + y -2 = 0
⇒ x + y = 2
Dividing by \(\sqrt{1+1}\) = √2 on both sides
(vi) √3x + y + 10 = 0
Answer:
(a) Slope – intercept form : y = -√3x – 10
Slope = – √3 , Y – intercept = – 10
(b) Intercept form: √3 x + y = – 10
(c) Normal form: √3 x + y = – 10
Dividing by \(\sqrt{3+1}\) = 2 both sides
Question 4.
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. (S.A.Q.)
Answer:
Equation of the lines x tan α + y sec α = 1
⇒ \(\frac{x}{\cot \alpha}+\frac{y}{\cos \alpha}\) = 1
This is in intercepts form and
a = cot α, b = cos α
given ab = product of intercepts on the axes = sin α
∴ ab = sin α
⇒ cot α . cos α = sin α
⇒ \(\frac{\cos \alpha}{\sin \alpha}\) .cos α = sin α
⇒ cos2 α = sin2 α
⇒ tan2 α = 1 ⇒ tan α = ± 1
(∵ 0 ≤ α ≤ \(\frac{\pi}{2}\)) we take α = \(\frac{\pi}{4}\) = 45°
Question 5.
If the sum of the reciprocals of the intercepts made by a variable straight line on the axes of coordinates is a constant, then prove that the line always passes through a fixed point. (S.A.Q.)
Answer:
Equation of the line in the intercept form is
\(\frac{x}{a}+\frac{x}{b}+\frac{x}{c}\) = 1 ……………. (1)
Sum of the reciprocals of the intercepts
Question 6.
Line L has intercepts a and b on the axes of the coordinates. When the axes are rotated through a given angle, keeping the origin fixed, the same line L has intercepts p and q on the transformed axes. Prove that \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}\). (S.A.Q.)
Answer:
Equation of the line in the original system in the intercepts form is \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ \(\frac{x}{a}+\frac{y}{b}\) – 1 = 0
Length of the perpendicular from origin
= \(\frac{|0+0-1|}{\sqrt{\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}}}\) ………………. (1)
Equation of the line in the second system in the intercepts form is
Question 7.
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 \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\). (S.A.Q)
Answer:
III.
Question 1.
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. (S.A.Q.)
Answer:
We have symmetric form of the line as
\(\frac{x-x_1}{\cos \alpha}=\frac{y-y_1}{\sin \alpha}\) = r
∴ Coordinates of any point on the given line are (x1 + r cos α , y1 + r sin α)
Given α = 30°
⇒ cos α = cos 30° = \(\frac{\sqrt{3}}{2}\)
⇒ sin α = sin 30° = \(\frac{1}{2}\)
∴ (x1, y1) = (- 2, 1)
⇒ x1 = – 2, y1 = 1
Taking r = 4, coordinates of P
= [- 2 + 4\(\left(\frac{\sqrt{3}}{2}\right)\), 1 + 4\(\left(\frac{1}{2}\right)\)]
= [- 2 + 2√3, 3]
Taking r = – 4 coordinates of P’ are
= [- 2 + 4\(\left(\frac{\sqrt{3}}{2}\right)\), 1 – 4\(\left(\frac{1}{2}\right)\)]
= (- 2 + 2√3, – 1)
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 ). (S.A.Q.)
Answer:
Equation of the line in symmetric form is
\(\frac{x-3}{\cos \alpha}=\frac{y-2}{\sin \alpha}\) = r
Coordinates of the point P are
= ( 3 + r cos α, 2 + r sin α)
= ( 3 + 5 cos α, 2 + 5 sin α)
P is a point on the line 3x – 4y – 1 = 0
⇒ 3 (3 + 5 cos α) – 4(2 + 5 sin α) – 1 = 0
⇒ 15 cos α – 20 sin α = 0
⇒ 15 cos α = 20 sin α
⇒ tan α = \(\frac{15}{20}\) = \(\frac{3}{4}\)
Case (i): When α is in the first quadrant
cos α = \(\frac{-4}{5}\), sin α = – \(\frac{3}{5}\)
Case (ii): When α is in the third quadrant then
cos α = \(\frac{-4}{5}\), sin α = – \(\frac{3}{5}\)
From case (i) coordinates of P are
[3 + 5\(\left(\frac{4}{5}\right)\), 2 + 5\(\left(\frac{3}{5}\right)\)]
Case (iii): Coordinates of P are
[3 – 5\(\left(\frac{4}{5}\right)\), 2 – 5\(\left(\frac{3}{5}\right)\)] = [- 1, 1]
Question 3.
A straight line whose inclination with the positive direction of the X-axis measured in the anti clockwise sense is \(\frac{\pi}{3}\) makes positive intercept on the Y – axis. If the straight line is at a distance of 4 from the origin, find its equation. (S.A.Q.)
Answer:
Given α = \(\frac{\pi}{3}\), p = 4
We have m = tan α = tan 60° = √3
∴ Equation of the line in slope – intercept form is y = √3 x + c
⇒ √3x – y + c = 0
Distance from the origin = 4
∴ \(\frac{|0-0+c|}{\sqrt{3+1}}\) = 4
⇒ |c| = 2 × 4 = 8
⇒ c = ±8
When c > 0 we have c = 8
Hence the equation of the line is
√3x – y + 8 = 0
Question 4.
A straight line L is drawn through the point A (2, 1) such that its point of intersection with the straight line x + y = 9 is at a distance
of 3√2 from A. Find the angle which the line L makes with the positive direction of the X-axis. (S.A.Q.)
Answer:
Suppose a is the angle made by L with the positive X-axis.
Any point on the line is (x1 + r cos α, y1 + r sin α)
= (2 + 3√2 cos α, 1 + 3√2 sin α)
This is a point over the line x + y = 9
∴ (2 + 3√2 cos α) + ( 1 + 3√2 sin α) = 9
⇒ 3 √2 ( cos α + sin α) = 6
⇒ cos α + sin α = \(\frac{6}{3 \sqrt{2}}\) = √2
⇒ \(\frac{1}{\sqrt{2}}\) cos α + \(\frac{1}{\sqrt{2}}\) sin α = 1
⇒ cos α . cos 45° + sin α . sin 45° = 1
⇒ cos (α – 45°) = cos 0°
⇒ α – 45° = 0 ⇒ α = 45° = \(\frac{\pi}{4}\)
Question 5.
A straight line L with negative slope passes through the point ( 8, 2 ) and cuts positive coordinate axes at the points P and Q. Find the minimum value of OP + OQ as L varies, where O is the origin. (S.A.Q.)
Answer:
Equation of the line passing through A (8, 2) with negative slope – m is
y – 2 = – m(x – 8)
⇒ mx + y – (2 + 8m) = 0
⇒ mx + y = 2 + 8m
∴ Minimum value of OP + OQ as L varies where O is the origin is 18.