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(x_{1}, y_{1}) 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(x_{1}, y_{1}), P(h, k)

‘P’ lies in ax + by + c = 0 then

ah + bk + c = 0

ah + bk = -c

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

Let the given point A(x_{1}, y_{1}) = (-1, 3)

Let (h, k) is the foot perpendicular from the point A(-1, 3) on line 5x – y – 18 = 0

∴ Foot of the perpendicular P(h, k) = (4, 2)

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(x_{1}, y_{1}), 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

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

Let the given point A(x_{1}, y_{1}) = (1, -2)

Now, B(h, k) be the image of A(1, -2) with respect to the straight line 2x – 3y + 5 = 0

\(\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)

∴ 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)

Let the slope of the required line is m_{2} = m

∴ The equation of the required line is passed through (-3, 2), and having slope m is y – y_{1} = m(x – x_{1})

y – 2 = m(x + 3) …….(2)

y – 2 = mx + 3m

mx – y + (3m + 2) = 0

Given that the angle between the lines is 45°,

Squaring on both sides

⇒ 10(m^{2} + 1) = 2(9m^{2} + 6m + 1)

⇒ 10m^{2} + 10 = 18m^{2} + 12m + 2

⇒ 8m^{2} + 12m – 8 = 0

⇒ 2m^{2} + 3m – 2 = 0

⇒ 2m^{2} + 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

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 L_{1} + λL_{2} = 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.

squaring on both sides

(-λ + 4)^{2} = 29λ^{2} + 32λ + 13

λ^{2} + 16 – 8λ = 29λ^{2} + 32λ + 13

28λ^{2} + 40λ – 3 = 0

28λ^{2} + 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 \(\overline{\mathbf{A D}}\).

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

⇒ SA^{2} = SB^{2}

⇒ (α + 2)^{2} + (β – 3)^{2} = (α – 2)^{2} + (β + 1)^{2}

⇒ α^{2} + 4 + 4α + β^{2} + 9 – 6β = α^{2} + 4 – 4α + β^{2} + 1 + 2β

⇒ 4α + 9 – 6β + 4α – 1 – 2β = 0

⇒ 8α – 8β + 8 = 0

⇒ α – β + 1 = 0 …….(1)

Also SB = SC

⇒ SB^{2} = SC^{2}

⇒ (α – 2)^{2} + (β + 1)^{2} = (α – 4)^{2} + (β – 0)^{2}

⇒ α^{2} + 4 – 4α + β^{2} + 2β + 1 = α^{2} + 16 – 8α + β^{2}

⇒ 4 – 4α + 2β + 1 – 16 + 8α = 0

⇒ 4α + 2β – 11 = 0 ……(2)

Solving (1) & (2)

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:

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)

Let S(α, β) be the circumcentre then SA = SB = SC

Now, SA = SB

⇒ SA^{2} = SB^{2}

⇒ (α – 1)^{2} + (β + 2)^{2} = (α – 2)^{2} + (β – 1)^{2}

⇒ α^{2} – 2α + 1 + β^{2} + 4 + 4β = α^{2} + 4 – 4α + β^{2} + 1 – 2β

⇒ -2α + 4β + 4α + 2β = 0

⇒ 2α + 6β = 0

⇒ α + 3β = 0 ……..(4)

Also SB = SC

⇒ SB^{2} = SC^{2}

⇒ (α – 2)^{2} + (β – 1)^{2} = (α + 2)^{2} + (β – 3)^{2}

⇒ α^{2} – 2α + 1 + β^{2} – 2β + 1 = α^{2} + 4α + 4 + β^{2} – 6β + 9

⇒ 4α + 9 – 6β + 4α – 1 + 2β = 0

⇒ 8α – 4β – 8 = 0

⇒ 2α – β + 2 = 0 ……..(5)

Solving (1) & (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)

Let the given point A(x_{1}, y_{1}) = (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 – y_{1} = m(x – x_{1})

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)

Squaring on both sides

⇒ m^{2} + 1 = (√3m – 1)^{2}

⇒ m^{2} + 1 = 3m^{2} + 1 – 2√3m

⇒ 2m^{2} – 2√3m = 0

⇒ m^{2} – √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 4p^{2} + q^{2} = a^{2}. [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(x_{1}, y_{1}) = (0, 0)

Now, p = the perpendicular distance from the point p(0, 0) to the straight line (1)

q = the perpendicular distance from the point p(0, 0) to the straight line (2)

L.H.S = 4p^{2} + q^{2}

= 4(a sin α cos α)^{2} + (a cos 2α)^{2}

= 4a^{2} sin^{2}α cos^{2}α + a^{2} cos^{2} 2α

= a^{2}(2 sin α cos α)^{2} + a^{2} cos^{2} 2α

= a^{2} sin^{2} 2α + a^{2} cos^{2} 2α

= a^{2}(sin^{2} 2α + cos^{2} 2α)

= a^{2}

= R.H.S

∴ 4p^{2} + q^{2} = a^{2}

### 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 – y_{1} = m(x – x_{1})

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.

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

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 – y_{1} = \(\frac{1}{m}\) (x – x_{1})

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 \(\sqrt{a^2+b^2}\)

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 + c_{1} = 0, we get

a = 10, b = -6, c_{1} = -8

Comparing (2) with ax + by + c_{2} = 0, we get

a = 10, b = -6, c_{2} = -9

Distance between the parallel lines =

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 m_{1} = \(\frac{-3}{7}\)

Slope of the line (2) is m_{2} = \(\frac{-7}{-p}=\frac{7}{p}\)

Since the given lines are perpendicular then,

m_{1} × m_{2} = -1

\(\frac{-3}{7} \times \frac{7}{p}\) = -1

p = 3

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)

∴ 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)

∴ 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)

∴ 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

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

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, \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\)

\(\frac{2}{b}-\frac{1}{a}=\frac{1}{c}\)

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 L_{1} + λL_{2} = 0

∴ \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a set of lines passing through the point of intersection of

L_{1} = x – 1 = 0 ……..(1)

L_{2} = 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.

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 L_{1} + λL_{2} = 0

∴ ax + by + c = 0 represents a set of lines passing through the point of intersection of

L_{1} = x – 1 = 0 …….(1)

L_{2} = 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(x_{1}, y_{1}) = (2, 3)

∴ The coordinates of P = (x_{1} + r cos θ, y_{1} + 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

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(x_{1}, y_{1}) = (-3, 5)

∴ The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope ‘1’ and passes through the point Q(-3, 5) is (y – y_{1}) = m(x – x_{1})

y – 5 = 1(x + 3)

y – 5 = x + 3

x – y + 8 = 0 …….(2)

Solving (1) & (2)

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(x_{1}, y_{1}) = (2, 3)

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 – y_{1} = m(x – x_{1})

y – 3 = 1(x – 2)

y – 3 = x – 2

x – y + 1 = 0 …….(2)

Solving (1) & (2)

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(x_{1}, y_{1}) = (4, 1)

Let (h, k) is the foot perpendicular from the point A(4, 1) on the line 3x – 4y + 12 = 0

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(x_{1}, y_{1}) = (3, 0)

Let (h, k) is the foot of the perpendicular from the point A(3, 0) on the line 5x + 12y – 41 = 0

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(x_{1}, y_{1}) = (1, 2)

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(x_{1}, y_{1}) = (-1, -3)

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.

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(x_{1}, y_{1}) = (3, -4)

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.

∴ 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)

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)

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

Now, SA = SB

⇒ SA^{2} = SB^{2}

⇒ (α – 1)^{2} + (β – 3)^{2} = (α – 0)^{2} + (β + 2)^{2}

⇒ α^{2} – 2α + 1 + β^{2} + 9 – 6β = α^{2} + β^{2} + 4 + 4β

⇒ 4 + 4β + 2α – 9 + 6β – 1 = 0

⇒ 2α + 10β – 6 = 0

⇒ α + 5β – 3 = 0 ……(1)

Also SB = SC

⇒ SB^{2} = SC^{2}

⇒ (α – 0)^{2} + (β + 2)^{2} = (α + 3)^{2} + (β – 1)^{2}

⇒ α^{2} + β^{2} + 4 + 4β = α^{2} + 9 + 6α + β^{2} + 1 – 2β

⇒ 9 + 6α – 2β + 1 – 4 – 4β = 0

⇒ 6α – 6β + 6 = 0

⇒ α – β + 1 = 0 ………(2)

Solving (1) & (2)

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.

Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC

Now, SA = SB

⇒ SA^{2} = SB^{2}

⇒ (α – 1)^{2} + (β – 3)^{2} = (α + 3)^{2} + (β – 5)^{2}

⇒ α^{2} – 2α + 1 + β^{2} + 9 – 6β = α^{2} + 9 + 6α + β^{2} + 25 – 10β

⇒ 6α + 25 – 10β – 1 + 2α + 6β = 0

⇒ 8α – 4β + 24 = 0

⇒ 2α – β + 6 = 0 ……(1)

Also SB = SC

⇒ SB^{2} = SC^{2}

⇒ (α + 3)^{2} + (β – 5)^{2} = (α – 5)^{2} + (β + 1)^{2}

⇒ α^{2} + 9 + 6α + 25 + β^{2} – 10β = α^{2} + 25 – 10α + β^{2} + 1 + 2β

⇒ 9 + 6α – 10β + 10α – 1 – 2β = 0

⇒ 16α – 12β + 8 = 0

⇒ 4α – 3β + 2 = 0 ………(2)

Solving (1) & (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.

Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC

Now, SA = SB

⇒ SA^{2} = SB^{2}

⇒ (α – 1)^{2} + (β – 0)^{2} = (α + 1)^{2} + (β – 2)^{2}

⇒ α^{2} – 2α + 1 + β^{2} = α^{2} + 2α + 1 + β^{2} + 4 – 4β

⇒ 2α – 4β + 4 + 1 – 1 + 2α = 0

⇒ 4α – 4β + 4 = 0

⇒ α – β + 1 = 0 ……….(1)

Also SB = SC

⇒ SB^{2} = SC^{2}

⇒ (α + 1)^{2} + (β – 2)^{2} = (α – 3)^{2} + (β – 2)^{2}

⇒ α^{2} + 2α + 1 + β^{2} + 4 – 4β = α^{2} + 9 – 6α + β^{2} + 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)

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)

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:

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)

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

∴ Vertex C = (-1, -3)

Let S(α, β) be the circumcentre of the triangle then, SA = SB = SC

Now, SA = SB

⇒ SA^{2} = SB^{2}

⇒ (α – 1)^{2} + (β + 1)^{2} = (α + 5)^{2} + (β – 5)^{2}

⇒ α^{2} – 2α + 1 + β^{2} + 1 + 2β = α^{2} + 25 + 10α + β^{2} + 25 – 10β

⇒ 50 + 10α – 10β – 1 + 2α – 2β – 1 = 0

⇒ 12α – 12β + 48 = 0

⇒ α – β + 4 = 0 ……..(4)

Also SB = SC

⇒ SB^{2} = SC^{2}

⇒ (α + 5)^{2} + (β – 5)^{2} = (α + 1)^{2} + (β + 3)^{2}

⇒ α^{2} + 25 + 10α + β^{2} + 25 – 10β = α^{2} + 2α + 1 + β^{2} + 6β + 9

⇒ 25 + 25 + 10α – 10β – 1 = -2α – 9 – 6β = 0

⇒ 8α – 16β + 40 = 0

⇒ α – 2β + 5 = 0 ……..(5)

Solving (4) & (5)

α = -3; β = 1

∴ Circumcentre S = (-3, 1)

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)

Let S(α, β) be the circumcentre of the triangle then, SA = SB = SC

Now, SA = SB

⇒ SA^{2} = SB^{2}

⇒ (α + 3)^{2} + (β – 1)^{2} = (α – 0)^{2} + (β + 2)^{2}

⇒ α^{2} + 9 + 6α + β^{2} + 1 – 2β = α^{2} + β^{2} + 4 + 4β

⇒ 6α – 2β + 9 + 1 – 4 – 4β = 0

⇒ 6α – 6β + 6 = 0

⇒ α – β + 1 = 0 ……..(4)

Also SB = SC

⇒ SB^{2} = SC^{2}

⇒ (α + 0)^{2} + (β + 2)^{2} = (α – 1)^{2} + (β – 3)^{2}

⇒ α^{2} + β^{2} + 4 + 4β = α^{2} – 2a + 1 + β^{2} – 6β + 9

⇒ 4 + 4β – 1 + 2α – 9 + 6β = 0

⇒ 2α + 10β – 6 = 0

⇒ α + 5β – 3 = 0 ……..(5)

Solving (4) & (5)

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)

Let the given point P(x_{1}, y_{1}) = (-10, 4)

Given that, tan θ = 2

cos θ = \(\frac{1}{\sqrt{5}}\)

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 – y_{1} = m(x – x_{1})

y – 4 = m(x + 10)

y – 4 = mx + 10m

mx – y + 10m + 4 = 0

If ‘θ’ be the angle between the lines (1) & (2)

Squaring on both sides

5(m^{2} + 1) = 5(m + 2)^{2}

5m^{2} + 5 = 5m^{2} + 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)

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 – y_{1} = m(x – x_{1})

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,

∴ The equations of the required straight lines are

from (2), y + 1 = 2 ± √3(x – 2).

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 – y_{1}) (x_{2} – x_{1}) = (x – x_{1})(y_{2} – y_{1})

⇒ (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

(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

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.

(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.

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(x_{1}, y_{1}) = (0, 0)

∴ The equation of the straight line passing through A(0, 0) and having slope ‘1’ is y – y_{1} = m(x – x_{1})

⇒ y – o = 1(x – 0)

⇒ y = x – o

⇒ y = x

⇒ x – y = 0

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(x_{1}, y_{1}) = (3, -2)

∴ The equation of the straight line passing through A(3, -2) and having slope ‘-1’ is y – y_{1} = m(x – x_{1})

⇒ 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 – y_{1}) (x_{2} – x_{1}) = (x – x_{1})(y_{2} – y_{1})

⇒ (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 – y_{1}) (x_{2} – x_{1})= (x – x_{1})(y_{2} – y_{1})

⇒ (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

The equation of the median through A is the equation of the straight line \(\overline{\mathrm{AD}}\),

(y – y_{1})(x_{2} – x_{1}) = (x – x_{1})(y_{2} – y_{1})

⇒ (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’:

= \(\frac{5}{12}\)

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 – y_{1} = m(x – x_{1})

⇒ y – 9 = \(\frac{-12}{5}\) (x + 4)

⇒ 5y – 45 = -12x – 48

⇒ 12x + 5y + 3 = 0

(iv) The equation of the perpendicular bisector of side \(\overline{\mathrm{AB}}\):

∴ 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 – y_{1} = m(x – x_{1})

⇒ 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

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)

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 = (x_{1} + |r| cos θ, y_{1} + |r| sin θ)

= (x_{1} ± r cos θ, y_{1} ± 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 = a^{2} + b^{2}

= 3^{2} + (-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.

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

⇒ cot^{2}α = 1

⇒ tan^{2}α = 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(x_{1}, y_{1}) = (-2, 1)

Inclination of straight line = 30°

distance |r| = 4

∴ Required points = (x_{1} + |r| cos θ, y_{1} + |r| sin θ)

= (x_{1} ± r cos θ, y_{1} ± 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

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

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(x_{1}, y_{1}) = (2, 3) and B(x_{2}, y_{2}) = (2, 10)

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(x_{1}, y_{1}) = (3, 2) and B(x_{2}, y_{2}) = (-4, -3)

Now, l_{11} = L(3, 2)

= 2(3) – 3(2) + 4

= 6 – 6 + 4

= 4 > 0

l_{22} = L(-4, -3)

= 2(-4) – 3(-3) + 4

= -8 + 9 + 4

= 5 > 0

Since l_{11} 0 and l_{22} > 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(x_{1}, y_{1}) = (2, -3) and B(x_{2}, y_{2}) = (3, -6)

(i) The X-axis divides the line segment AB in the ratio is,

-y_{1} : y_{2} = -(-3) : -6

= 3 : -6

= 1 : -2

= -1 : 2

(ii) Y-axis divides \(\overline{\mathrm{AB}}\) in the ratio is,

-x_{1} : x_{2} = -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)

∴ 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(x_{1}, y_{1}) = (2, -1) and B(x_{2}, y_{2}) = (1, 1)

Question 67.

Transform the equation (2 + 5k)x – 3(1 + 2k)y + (2 – k) = 0 into the form L_{1} + λL_{2} = 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 l_{1} + λl_{2} = 0

where l_{1} = 2x – 3y + 2 = 0 ……..(1)

l_{2} = 5x – 6y – 1 = 0 ……..(2)

Solving (1) & (2)

∴ Point of concurrence = (5, 4)

Question 68.

Transform the equation (k + 1)x + (k + 2)y + 5 = 0, into the form L_{1} + λL_{2} = 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 l_{1} + λl_{2} = 0

where l_{1} = x + 2y + 5 = 0 ………(1)

l_{2} = x + y = 0 ……..(2)

Solving (1) & (2)

∴ 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(x_{1}, y_{1}) = (-1, 2) & B(x_{2}, y_{2}) = (5, -1)

The equation of the straight line passing through the points A(-1, 2) & B(5, -1) is (y – y_{1})(x_{2} – x_{1}) = (x – x_{1})(y_{2} – y_{1})

⇒ (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(x_{1}, y_{1}) = (1, 2) & B(x_{2}, y_{2}) = (3, 4)

Given that, the points (1, 2) & (3, 4) lie on the same direction 3x – 5y + a = 0, then -l_{11} : l_{22} < 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

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 a_{1}x + b_{1}y + c_{1} = 0, we get

a_{1} = 2, b_{1} = 1, c_{1} = 4

Comparing (2) with a_{2}x + b_{2}y + c_{2} = 0, we get

a_{2} = 3, b_{2} = -1, c_{2} = 7

If ‘θ’ is the angle between the lines (1) & (2)

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 a_{1}x + b_{1}y + c_{1} = 0, we get

a_{1} = √3, b_{1} = 1, c_{1} = 1

Comparing (2) with a_{2}x + b_{2}y + c_{2} = 0, we get

a_{2} = 1, b_{2} = 0, c_{2} = 1

If ‘θ’ is the angle between the lines (1) & (2)

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 a_{1}x + b_{1}y + c_{1} = 0, we get

a_{1} = a, b_{1} = b, c_{1} = -a – b

Comparing (2) with a_{2}x + b_{2}y + c_{2} = 0, we get

a_{2} = a + b, b_{2} = b – a, c_{2} = -2b

If ‘θ’ is the angle between the lines (1) & (2)

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

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(x_{1}, y_{1}) = (4, -3)

Equation of the straight line passing through (4, -3) and having slope \(\frac{-3}{5}\) is,

y – y_{1} = \(\frac{-1}{m}\)(x – x_{1})

⇒ 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 m_{1} = m_{2}

⇒ \(\frac{3}{5}\) = \(\frac{k}{5}\)

⇒ 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).

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).

The equation of straight line passing through E & slope -3 is y – y_{1} = \(\frac{-1}{m}\)(x – x_{1})

⇒ y – 2 = -3(x – 2)

⇒ y – 2 = -3x + 6

⇒ 3x + y – 8 = 0

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)

Let ‘A’ be the angle between the lines (1) & (3) then,

∴ 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)

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)

Question 83.

Find the incentre of the triangle whose vertices are (1, √3), (2, 0), and (0, 0).

Solution:

Given, A (x_{1}, y_{1}) = (1, √3), B(x_{2}, y_{2}) = (2, 0), C(x_{3}, y_{3}) = (0, 0)

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)

Vertex A:

Solving (1) & (3)

from (1), x = 1

from (3), x + y = 1

⇒ 1 + y = 1

⇒ y = 0

∴ Vertex A = (1, 0)

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)

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 a_{1}x + b_{1}y + c_{1} = 0, we get

a_{1} = k, b_{1} = 1, c_{1} = 9

Comparing (2) with a_{2}x + b_{2}y + c_{2} = 0, we get

a_{2} = 3, b_{2} = -1, c_{2} = 4

Given that, θ = \(\frac{\pi}{4}\)

Let ‘θ’ is the angle between the lines (1) & (2) then

Squaring on both sides

⇒ (k^{2} + 1)(10) = 2(3k – 1)^{2}

⇒ 10k^{2} + 10 = 18k^{2} + 2 – 12k

⇒ 8k^{2} – 12k – 8 = 0

⇒ 2k^{2} – 3k – 2 = 0

⇒ 2k^{2} – 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)

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 – y_{1})(x_{2} – x_{1}) = (x – x_{1})(y_{2} – y_{1})

⇒ (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)

∴ 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 – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

⇒ y – 1 = \(\frac{3}{2}\) (x + 2)

⇒ 2y – 2 = 3x + 6

⇒ 3x – 2y + 8 = 0

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

The perpendicular distance from the point Q(-4, a) to the straight line 3x + 4y – 8 = 0 is,

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)

⇒ -6y – 8 = 21x – 35

⇒ 21x + 6y – 27 = 0

⇒ 7x + 2y – 9 = 0

The equation of the diagonal \(\overline{\mathrm{BD}}\) is,

(y – y_{1})(x_{2} – x_{1}) = (x – x_{1})(y_{2} – y_{1})

⇒ (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)

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)

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)

∴ 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 – y_{1} = m(x – x_{1})

⇒ \(y-\frac{-3}{5}=\frac{-3}{4}\left(x-\frac{21}{5}\right)\)

⇒ \(\frac{5 y-3}{5}=\frac{-3}{4}\left(\frac{5 x-21}{5}\right)\)

⇒ 5y – 3 = \(\frac{-3}{4}\)(5x – 21)

⇒ 20y – 12 = -15x + 63

⇒ 15x + 20y – 75 = 0

⇒ 3x + 4y – 15 = 0

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 m_{1} = \(\frac{-3}{p}\)

Slope of (2) is m_{2} = \(\frac{-7}{-3}\) = \(\frac{7}{3}\)

Since (1) & (2) are perpendicular then

m_{1} × m_{2} = -1

⇒ \(\frac{-3}{p} \times \frac{7}{3}=-1\)

⇒ \(\frac{-7}{p}\) = -1

⇒ p = 7