## Rajasthan Board RBSE Class 11 Maths Chapter 11 Straight Line Ex 11.3

Question 1.

Find the angle between the following straight lines :

(i) y = (2 – \(\sqrt { 3 }\)) x + 5 and y = (2 + \(\sqrt { 3 }\)) x – 7

(ii) 2y – 3x + 5 = 0 and 4x + 5y + 8 = 0

(iii) \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1 and \(\frac { x }{ b } \) – \(\frac { y }{ a } \) = 1

Solution:

(i) y = (2 – \(\sqrt { 3 }\)) x + 5

Comparing y = m_{1} x + c_{1}

m_{1} = 2 – \(\sqrt { 3 }\)

and y = (2 + \(\sqrt { 3 }\))x – 7

Comparing y = m_{2}x + c_{2}

m_{2} = 2 + \(\sqrt { 3 }\)

Angle between both lines

From equation (i)

m_{1} = tan θ_{1} = \(\frac { b }{ a } \)

From equation (ii)

m_{2} = tan θ_{2} = \(\frac { a }{ b } \)

Therefore, angle between both lines is

Question 2.

Prove that following straight lines are parallel

(i) 2y = mx + c and 4y = 2mx

(ii) x cos α + y sin α = p and x + y tan α = 5 tan α

Solution:

(i) 2y = mx + c

Comparing with y = m_{1}x + c_{1}

m_{1} = \(\frac { m }{ 2 } \)

4y = 2 mx

Thus, both lines will be parallel.

(ii) x cos α + y sin α = p

⇒ y sin α = – x cos α + p

We see that

m_{1} = m_{2} = – cot α

Therefore both line will be parallel.

Question 3.

Prove that lines whose equation are 4x + 5y + 7 = 0 and 5x – 4y – 11 = 0, are perpendicular to each other.

Solution:

4x + 5y + 7 = 0

⇒ 5y = -4x – 7

Therefore both lines will be perpendicular.

Question 4.

Find the equations of those straight lines which

(i) Passing through point (4,5) and is parallel to line 2x – 3y – 5 = 0.

(ii) Passing through point (1,2) and is perpendicular to line 4x + 3y + 8 = 0.

(iii) Is parallel to line 2x + 5y = 7 and passes through the mid point of line joining the points (2,7) and (-4,1).

(iv) Divide the line joining the points (- 3, 7) and (5, – 4) in the ratio 4 : 7 and is perpendicular to this.

Solution:

(i) Equation of line parallel to line

2x – 3y – 5 = 0

2x – 2y + c = 0

It passes through point (4,5)

then 2 × 4 -3 × 5 + c = 0

⇒ 8 – 15 + c = 0

⇒ c = 7

Hence, equation of straight line is

2x – 3y + 7 = 0

(ii) Equation of line perpendicular to line

4x + 3y + 8 = 0

3x – 4y + c = 0

This line passes through point (1,2), then

3 × 1 – 4 × 2 + c = 0

⇒ 3 – 8 + c = 0

⇒ c = 5

Thus, equation to straight line is

3x – 4y + 5 = 0

(iii) Equation of line parallel to line

2x + 5y = 7

2x + 5y = c …(i)

Mid point of line joining the points (2,7) and (-4, 1)

Putting this value in equation (i),

⇒ 2 × (-1) + 5 × (4) = c

⇒ -2 + 20 = c

⇒ c = 18

Thus, Required equation of line is

2x + 5y = 18

(iv) Equation of line passing through the points (- 3, 7) and (5, – 4)

Equation of line perpendicular to this line

11y – 8x =c …(i)

Coordinates of point which divide the line joining the points (-3, 7) and (5, -4) in the ratio 4 : 7

Line passes through this point so putting this point in equation (i)

Thus, equation of required line

11y – 8x = \(\frac { 371 }{ 11 } \)

⇒ 121y – 88x = 371

⇒ 88x – 121y + 371 = 0

Question 5.

The vertices of a triangle are (0, 0), (4, – 6) and (1, – 3) Find the equations of perpendiculars drawn from these points to the corresponding sides.

Solution:

Let altitudes of ∆ABC are AD, BE, CF then AD ⊥BC, BE ⊥ CA and CF ⊥ AB Now slope of side BC

Equation of perpendicular AD Which passes through point (0, 0)

y – 0 = 1(x – 0)

⇒ y = x

⇒ y – x = 0

Equation of perpendicular BE, which passes through point (4, – 6)

y + 6= \(\frac { 1 }{ 3 } \) (x – 4)

⇒ 3y + 18 – x + 4 = 0

⇒ 3y – x + 22 = 0

⇒ x – 3y = 22

Equation of perpendicular CF which passes through point (1,-3)

y + 3 = \(\frac { 2 }{ 3 } \) (x – 1)

⇒ 3y + 9 – 2x + 2 = 0

⇒ 2x – 3y = 11

Question 6.

Find the ortho center of triangle whose vertices are (2, 0), (3, 4) and (0, 3).

Solution:

Let vertices of triangle are A(2, 0), B(3, 4) and C(0, 3)

Multiply equation (i) by 4 and subtracting this from equation (ii).

Question 7.

Two vertices of a triangle are (3, -1) and (- 2. 3). Orthocentre of triangle lies at the origin. Find the coordinates of third vertex.

Solution:

Vertices of ∆ABC are A(h, k), 5(3, – 1) and (- 2,3). vertex A is intersecting point of AB and AC. Side AB passes through point (3, – 1) and is perpendicular to DC whose slope is – \(\frac { 3 }{ 2 } \)

Similarly AC passes through (- 2,3) and is perpendicular to OB, whose slope is – \(\frac { 1 }{ 3 } \). Thus equation of side AC

y – 3 = 3(x + 2)

⇒ 3x – y + 9 = 0 …(ii)

Solving equation (i) and (ii)

x = – \(\frac { 36 }{ 7 } \) and y = – \(\frac { 45 }{ 7 } \)

Thus, coordinates of A is ( – \(\frac { 36 }{ 7 } \) – \(\frac { 45 }{ 7 } \))

Question 8.

Find the equation of perpendicular bisector qf line joining the points (2, -3) and (- 1, 5).

Solution:

AB is a line joining the points A(2, -3) and B(-1, 5)

Equation of line perpendicular to this line

Question 9.

Find the equation of line which is perpendicular on straight line \(\frac { x }{ a } \) – \(\frac { y }{ b } \) = 1 at the point where it meets the x-axis

Solution :

Let equation of line AB is \(\frac { x }{ a } \) – \(\frac { y }{ b } \) =1, then co-ordinates of that point where it meets with x-axis will be (a, 0). We get equation of line ⊥ to this line.

Question 10.

Find the equation of the line which is parallel to the line 2x + 3y + 11 = 0 and sum of intercepts cut by axis is 15.

Solution:

Equation of given line

2x + 3y + 11 = 0

Equation of line parallel to this line.

2x + 3y + c = 0 ….(i)

⇒ 2x + 3y = -c

Question 11.

Find the equation of straight lines which passes through the point (2, – 3) and makes an angle of 45° with line 3x – 2y = 4.

Solution:

Equation of line passing through point (2, -3)

y + 3 = m_{1} (x – 2)

⇒ y = m_{1} (x – 2) -3

⇒ y = m_{1} x – (2m_{1} + 3)

Eq^{n} of given straight line

3x – 2y = 4

⇒ 2y = 3x – 4

Thus, equation of line

y + 3 = – 5 (x – 2)

⇒ y + 3 + 5(x – 2) = 0

⇒ y + 5x – 7 = 0

Thus, equation of line

y + 3 = \(\frac { 1 }{ 5 } \) (x – 2)

⇒ 5y + 15 – x + 2 = 0

⇒ 5y – x + 17 = 0

Thus, equation of lines are

y + 5x – 7 = 0 and 5y – x + 17 = 0

Question 12.

Find the equation of straight line which passes through the point (4, 5) and make similar angles with lines 3x = 4y + 7 and 5y = 12x + 6

Solution:

Equation of line passing through the point (4, 5)

y – 5 = m_{1} (x – 4) …(i)

Given line 3x = 4y + 7

⇒ 4y = 3x – 7

⇒ y = \(\frac { 3 }{ 4 } \) x – \(\frac { 7 }{ 4 } \) …(ii)

Comparing with y = m_{2} x + c_{2}

m_{2} = \(\frac { 3 }{ 4 } \)

other line 5y = 12x + 6

⇒ y = \(\frac { 12 }{ 5 } \) x + \(\frac { 6 }{ 5 } \) …(iii)

Comparing with y = m_{3}x + c_{3}

According to question m_{3} = \(\frac { 12 }{ 5 } \)

Angle between line (i) and (ii) = Angle between line (i) and (iii)

Question 13.

Prove that following will be equation of line which passes through origin and makes angle θ with line y = mx + c

Solution:

Let equation of line PQ is y = mx + c and it makes angle θ with PQ and its gradient = m_{1
}

Question 14.

Prove that equation of line which passes through a point (a cos^{3} θ, a sin^{3} θ) and is perpendicular to the x sec θ + y cosec θ = a is x cos θ – y sin θ = a cos 2θ

Solution:

Line x sec θ + y cosec θ = a

⇒ y sin θ – a sin^{4} θ = x cos θ – a cos^{4} θ

⇒ x cos θ – y sin θ = a cos^{4} θ – a sin^{4} θ

⇒ x cos θ – y sin θ = a (cos^{4} θ – sin^{4} θ)

⇒ xcos θ – y sin θ = a (cos^{2} θ + sin^{2} θ)

(cos^{2} θ – sin^{2} θ)

⇒ x cos θ – y sin θ = a × 1 × cos2θ

⇒ x cos θ – y sin θ = a cos2θ

Thus equation of required perpendicular line

x cos θ – y sin θ = a cos2θ

Question 15.

Vertex of an equilateral triangle is (2, 3) and equation of its opposite side is x + y = 2. Find the equation of the remaining sides.

Solution:

Let, In ∆ABC, coordinate of A are (2,3) and equation of BC is x + y = 2. Equation of line passes through point A (2,3) and making angle of 60° with x + y = 2 is.

Question 16.

Find the equation oftwo lines which pass through point (3,2) and makes an angle of 60° with line x + \(\sqrt { 3 }\)y = 1.

Solution:

We know that equation of a line passing through point (h, k) and makes angle α with line y = mx + c is

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