## 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 = m_{1}x + c_{1}

m_{1} = 0

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

m_{2} = 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 – y_{1} = m (x – x_{1})

⇒ 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

m_{1} = m_{2}

⇒ 2m – 2

⇒ m = 1

(ii) lines will be perpendicular if

m_{1}m_{2} = -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

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