RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Ex 11.3

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 = m₁ x + c₁
m₁ = 2 – \(\sqrt { 3 }\)
and y = (2 + \(\sqrt { 3 }\))x – 7
Comparing y = m₂x + c₂
m₂ = 2 + \(\sqrt { 3 }\)
Angle between both lines


From equation (i)
m₁ = tan θ₁ = \(\frac { b }{ a } \)
From equation (ii)
m₂ = tan θ₂ = \(\frac { a }{ b } \)
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 = m₁x + c₁
m₁ = \(\frac { m }{ 2 } \)
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
m₁ = m₂ = – 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 = m₁ (x – 2)
⇒ y = m₁ (x – 2) -3
⇒ y = m₁ x – (2m₁ + 3)
Eqⁿ 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 = m₁ (x – 4) …(i)
Given line 3x = 4y + 7
⇒ 4y = 3x – 7
⇒ y = \(\frac { 3 }{ 4 } \) x – \(\frac { 7 }{ 4 } \) …(ii)
Comparing with y = m₂ x + c₂
m₂ = \(\frac { 3 }{ 4 } \)
other line 5y = 12x + 6
⇒ y = \(\frac { 12 }{ 5 } \) x + \(\frac { 6 }{ 5 } \) …(iii)
Comparing with y = m₃x + c₃
According to question m₃ = \(\frac { 12 }{ 5 } \)
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 = m₁

Question 14.
Prove that equation of line which passes through a point (a cos³ θ, a sin³ 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 sin⁴ θ = x cos θ – a cos⁴ θ
⇒ x cos θ – y sin θ = a cos⁴ θ – a sin⁴ θ
⇒ x cos θ – y sin θ = a (cos⁴ θ – sin⁴ θ)
⇒ xcos θ – y sin θ = a (cos² θ + sin² θ)
(cos² θ – sin² 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

RBSE Solutions for Class 11 Maths