RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise

Rajasthan Board RBSE Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise

Question 1.
AB is a line joining the points A(2, -3) and B(-1, 5) Equation of line perpendicular to this line
(A) y = 5
(B) x = 5
(C) x = -5
(D) y = -5
Solution:
Equation of line parallel to y-axis is x = -5

Thus, option (c) is correct.

Question 2.
Equation of a line which passes through point (3, – 4) and is parallel to x-axis is :
(A) x – 3
(B) y = -4
(C) x + 3 = 0
(D) y – 4 = 0
Solution:
Equation of line parallel to x-axis is y = -4

Thus, option (b) is correct.

Question 3.
Slope of y-axis is :
(A) 1
(B) 0
(C) ∞
(D) π/2
Solution:
Slope y – axis = tan 90° = ∞

Thus, option (c) is correct.

Question 4.
Line represented by equation x × \(\frac { 1 }{ 2 } \) + y × \(\frac { \sqrt { 3 } }{ 2 } \) = 5 is in following form:
(A) Symmetrical form
(B) Slope form
(C) Intercept form
(D) Normal form
Solution:

Question 5.
Equation of line passsing through origin and parallel to line 3x – 4y = 7 is :
(A) 3x – 4y = 1
(B) 3x – 4y – 0
(C)4x – 3y = 1
(D) 3y – 4x = 0
Solution:
Equation of line parallel to straight line
3x – 4y = 7
3x – 4y = c
Since it passes through origin
∴ 3 (0) – 4 (0) = c
⇒ c = 0
Thus, Required equation is 3x – 4y = 0
Therefore, option (b) is correct.

Question 6.
It is the length of perpendicular drawn from origin to the line x + \(\sqrt { 3 }\)y = 1 then value of p is :

Solution:

⇒ x cos 60° + y sin 60° = \(\frac { 1 }{ 2 } \)
Comparing (i) by x cos α + y sin α = p
p = \(\frac { 1 }{ 2 } \)
Thus, option (B) is correct.

Question 7.
If lines y = mx + 5 and 3x + 5y = 8 are perpendicular to each other, then value of m is :

Solution:
y = mx + 5 …(i)
and 3x + 5y = 8
⇒ 5y = -3x + 8
⇒ y = \(\frac { -3 }{ 5 } \)x + 8 …(ii)
Lines (i) and (ii) are perpendicular to each other

Thus, option (a) is correct.

Question 8.
Equation of line passes through point (1, – 2) and perpendicular to line 3x – 4y + 7 = 0 will be :
(A) 4x + 3y – 2 = 0
(B) 4x + 3y + 2 = 0
(C) 4x – 3y + 2 = 0
(D) 4x – 3y – 2 = 0
Solution:
Equation of line perpendicular to
3x – 4y + 7 = 0
4x + 3y + c = 0
It passes through point (1,2)
4(1) + 3(-2) + c = 0
⇒ 4 – 6 + c = 0
⇒ c – 2=0
⇒ c = 2
Thus, equation is 4x + 3y + 2 =0
Thus, option (b) is correct.

Question 9.
Obtuse angle between lines y = -2 and y = x + 2 is :
(A) 145°
(B) 150°
(C) 135°
(D) 120°
Solution:
comparing y = -2x with y = m₁x + c₁
m₁ = 0
comparing y = x + 2 with y = m₂x + c₂
m₂ = 1
Angle between both lines

When tan θ = +1 then θ = 45°
when tan θ = -1 then θ = 135°
Thus, option (c) is correct

Question 10.
Length of intercepts cut at x and y axis by line 3x – 4y – 4 = 0 is :

Solution:

Thus, option (a) is correct

Question 11.
Line joining the points (1, 0) and (-2, \(\sqrt { 3 }\) ) makes an angle θ with x-axis then value of tan θ is :

Solution:
Slope of line joining the points (1,0) and (-2, \(\sqrt { 3 }\))
m = tanθ

Thus, option (d) is correct.

Question 12.
By reducing the equation of line 2x + \(\sqrt { 3 }\) y – 4 = 0 in slope form, constant used in slope form is :

Solution:

Thus, option (d) is correct.

Question 13.
Find the equation of line which passes through point (2, 3) and makes an angle of 45° with x- axis.
Solution:
Given θ = 45°
then tan θ = m = tan 45° = 1
⇒ m = 1
equation of line passing through point (2, 3)
y – y₁ = m (x – x₁)
⇒ y – 3 = 1 (x – 2)
⇒ y – 3 = x – 2
⇒ x – y + 3 – 2 = 0
⇒ x – y + 1 = 0
Thus required equation of straight line is
x – y + 1 = 0

Question 14.
Find the equation of line which passes through points (-3, 2) and cut equal and opposite sign intercepts with axis.
Solution:
Let a and – a are intercepts cut by line, then its equation

Question 15.
If length of perpendicular from origin to straight line 4x + 3y + a = 0 is 2, then find value of a.
Solution:
Given equation
4x + 3y + a = 0
Length of perpendicular drawn from origin (0,0) to line 4x + 3y + a = 0

Question 16.
If intercepts cut at axis by any line bisects at point (5, 2), then find the equation of line.
Solution:
Let equation of line is \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1 which meets x and y axis at points A(a, 0) and 5(0, b) coordinates of mid point of AB are

But it is given that mid point of AB is a
⇒ \(\frac { a }{ 2 } \) = 5
⇒ a = 10
⇒ \(\frac { b }{ 2 } \) = 2
⇒ b = 4
∴ Required equation
\(\frac { x }{ 10 } \) + \(\frac { y }{ 4 } \) = 1
⇒ 4x + 10y = 40
or 2x + 5y = 20

Question 17.
Find the equation of line which passes through a point (0,1) and intercept cut by line at axis as 3 times the intercept cut at y-axis.
Solution:
Let equation of line
\(\frac { x }{ 10 } \) + \(\frac { y }{ 10 } \) = 1 …(i)
Given that intercept of x-axis is 3 times the intercept of y-axis
a = 3b
Putting a = 3b in eq.ⁿ (i)
\(\frac { x }{ 3b } \) + \(\frac { y }{ b } \) = 1
⇒ x + 3y = 3b
But it passes through point (0, 1)
∴ 0 + 3 (1) = 3b
⇒ b = 1
∴ a = 3 (1) = 3
∴ a = 3, b = 1
From eq.ⁿ (ii) x + 3y = 3 × 1
Thus, required equation of line
⇒ x + 3y = 3

Question 18.
If straight lines y = 2mx + c and 2x – y + 5 – 0 are parallel and perpendicular to each other, then find the value of m.
Solution:
Given lines
y = 2mx + c …(i)
and 2x – y + 5 = 0
⇒ y = 2x + 5 …(ii)
From equations (i) and (ii)
(i) lines will be II if
m₁ = m₂
⇒ 2m – 2
⇒ m = 1
(ii) lines will be perpendicular if
m₁m₂ = -1
⇒ 2m × 2 = -1
⇒ m = \(\frac { -1 }{ 4 } \)
Thus values of m will be 1 and \(\frac { -1 }{ 4 } \)

Question 19.
If length of perpendicular from origin to straight line 4x + 3y + a = 0 is 2, then find the value of a.
Solution:
Given equation 4x + 3y + a = 0
Length of perpendicular drawn from origin (0, 0) to line 4x + 3y + a = 0

Question 20.
If length of perpendicular from origin to line \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1 is p, then prove that

Solution:
Equation of line cut intercepts a and b at axis
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1 …(i)
Length of perpendicular drawn from origin (0,0) to line

RBSE Solutions for Class 11 Maths