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

June 14, 2019 by Fazal Leave a Comment

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
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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° = ∞
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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:
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise

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 :
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
Solution:
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
⇒ 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 :
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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 :
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
Solution:
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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 :
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
Solution:
Slope of line joining the points (1,0) and (-2, \(\sqrt { 3 }\))
m = tanθ
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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 :
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
Solution:
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise

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

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

Question 20.
If length of perpendicular from origin to line \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1 is p, then prove that
RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise
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 Chapter 11 Straight Line Miscellaneous Exercise

RBSE Solutions for Class 11 Maths

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