## Rajasthan Board RBSE Class 11 Maths Chapter 11 Straight Line Ex 11.2

Question 1.

Convert the following equations in slope form and intercept form and find the value of constants used in standard form.

(i) 7x – 13y = 15

(ii) 5x + 6y + 8 = 0

Solution:

Question 2.

Find the slope of line x cos α + y sin α = p

Solution :

x cos α + y sin α =p

⇒ y sin α = – x cos α – p

Question 3.

Find the tangent of the following lines with +ve x-axis.

(i) x – y + 2 = 0

(ii) x + y – 2 = 0

Solution:

(i) x – y + 2 = 0

⇒ y = x + 2

Comparing with y = mx + c

m =

⇒ m = tan θ

tan θ = = tan 60°

Thus tangent of required angle is 60°.

Question 4.

Prove that (x_{1}, y_{1}) will be coordinate of mid point of section cut by line

Solution:

Let straight line cuts x and y axis of points B and A and coordinates of points A and B are (0, 2y_{1})

Question 5.

Find the length of intercept cut between the axis from straight line 3x + 4y = 6 and also find its mid point.

Solution:

3x + 4y = 6

⇒ + = 1

Comparing this equation + = 1, a = 2 and b =

Straight line intersect the axis at the points (0,) at and (2,0)

Question 6.

Find the values of a and b where equations 5x – 4y = 20 and ax – by + 1 = 0 represent same straight line.

Solution:

Question 7.

Reduce the following equations in the form x cos α + y sin α = p.

(i) x + y + = 0

(ii) x – y + 2 = 0

Solution:

(i) x + y + = 0

⇒ x + y = –

⇒ – x – y =

Dividing both side by

Thus perpendicular form of given equation.

x cos 225° + 7 sin 225° = 1

Thus perpendicular form of given equation

x cos 150° + y sin 150° = 1

Question 8.

Reduce the straight line 3x – 4y – 11 = 0 in the normal form and find length of perpendicular from origin to line and its slope from x-axis.

Solution:

3x – 4y – 11 = 0

3x – 4y = 11

Question 9.

+ = 1 and 2x – 3y = 5 represent same line, then find the values of a and b.

Solution :

Question 10.

If straight line y = mx + c and x cos α + 7 sin α = p represent same line, then find slope of line with x axis, and length of intercept cut from y – axis.

Solution:

y = mx+c …(i)

x cos α + 7 sin α = p

⇒ 7 sin α = -x cos α + p

Comparing equation (i) and (ii)

m = tan θ = -cot α

⇒ m = tan θ = tan (90° + α)

Thus θ = 90° + α

Slope of line = 90° + α

Thus slope of line is 90° + α and intercept is p cosec α.

Question 11.

Find the equation of straight line which passes through point (2, 3) and makes an angle of 45° with x- axis.

Solution :

Here point (x_{1},y_{1}) = (2,3) and slope θ = 45°

⇒ tan θ = tan 45° = 1

∴ Equation of line y – y_{1} = m (x – x_{1})

⇒ y – 3 = 1 (x – 2)

⇒ y – 3 = x – 2

⇒ x – y + 1 = 0

Question 12.

Find the equation of line passing through the following two given points :

(i) (3, 4) and (5, 6)

(ii) (0, – a) and (b, 0)

(iii) (a, b) and (a + b, a- b)

(iv) (at_{1}, alt_{1}) and (at_{2}, alt_{2})

(v) (a sec α, b tan α) and (α sec β, b tan β)

Solution:

(i) Equation of line passing through (3, 4) and (5, 6)

(ii) Equation of line passing through (0, – a) and (b, 0)

(iii) Equation of line passing through (a, b) and (a + b, a – b)

(iv) Equation of line passing through (at_{1} alt_{1}) and (at_{2}, alt_{2})

(v) Equation of line passing through (a sec α, b tan α) and (a sec β, b tan β)

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