Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.7

Question 1: Find the equation of a line for which

(i) p = 5, α = 60°

(ii) p = 4, α = 150°

(iii) p = 8, α = 225°

(iv) p = 8, α = 300°

Solution:

(i) p = 5, α = 60°

Given: p = 5, α = 60°

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x cos 60° + y sin 60° = 5

x/2 + √3y/2 = 5

x + √3y = 10

Therefore, the equation of line is x + √3y = 10.

(ii) p = 4, α = 150°

Given: p = 4, α = 150°

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x cos 150° + y sin 150° = 4

x cos(180° – 30°) + y sin(180° – 30°) = 4 [As, cos (180° – θ) = – cos θ , sin (180° – θ) = sin θ]

– x cos 30° + y sin 30° = 4

–√3x/2 + y/2 = 4

-√3x + y = 8

Therefore, the equation of line is -√3x + y = 8.

(iii) p = 8, α = 225°

Given: p = 8, α = 225°

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x cos 225° + y sin 225° = 8

– x/√2 – y/√2  = 8

x + y + 8√2  = 0

Therefore, the equation of line is x + y + 8√2  = 0

(iv) p = 8, α = 300°

Given: p = 8, α = 300°

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x cos 300° + y sin 300° = 8

x/2 – y√3/2 = 8

x – √3y = 16

Therefore, the equation of line is x – √3y = 16

Question 2: Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is 30°.

Solution:

Given: p = 4, α = 30°

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x cos 30° + y sin 30° = 4

x√3/2 + y1/2 = 4

√3x + y = 8

Therefore, the equation of line is √3x + y = 8.

Question 3: Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x-axis is 15°.

Solution:

Given: p = 4, α = 15°

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x cos 15° + y sin 15° = 4

Now as, cos 15° = cos (45° – 30°) = cos45°cos30° + sin45°sin30°  [Since, cos (A – B) = cos A cos B + sin A sin B ]

= 1/√2 × √3/2 + 1/√2 × 1/2

= 1/2√2( √3 + 1 )

And sin 15 = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30°  [Since, Sin (A – B) = sin A cos B – cos A sin B ]

= 1/√2 × √3/2 – 1/√2 × 1/2

= 1/2√2(√3 – 1)

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x × [1/2√2(√3 + 1)] + y × [1/2√2(√3 – 1)]

(√3+1)x +(√3-1) y = 8√2

Therefore, the equation of line is (√3+1)x +(√3-1) y = 8√2.

Question 4: Find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle α given by tan α = 5/12 with the positive direction of x–axis.

Solution:

Given: p = 3, α = tan^-1 (5/12)

tan α = 5/12

So,

sin α = 5/13

cos α = 12/13

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

12x/13 + 5y/13 = 3

12x + 5y = 39

Therefore, the equation of line is 12x + 5y = 39.

Question 5: Find the equation of the straight line on which the length of the perpendicular from the origin is 2 and the perpendicular makes an angle α with x–axis such that sin α = 1/3.

Solution:

Given: p = 2, sin α = 1/3

As, cos α = √(1 – sin² α)

= √(1 – 1/9)

= 2√2/3

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x2√2/3 + y/3 = 2

2√2x + y = 6

Therefore, the equation of line is 2√2x + y = 6.

Question 6: Find the equation of the straight line upon which the length of the perpendicular from the origin is 2 and the slope of this perpendicular is 5/12.

Solution:

Given: p = ±2

tan α = 5/12

Therefore, sin α = 5/13

cos α = 12/13

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

12x/13 + 5y/13 = ±2

12x + 5y = ±26

Therefore, the equation of line is 12x + 5y = ±26.

Question 7: The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150° with the positive direction of y-axis. Find the equation of the line.

Solution:

Given: p = 7 (perpendicular distance from origin)

Also given that the angle made with y-axis is 150°

therefore, the angle made with x-axis is 180° – 150° = 30°

sin 30° = 5/13

cos 30° = 12/13

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x(√3/2) + y(1/2) = 7

√3x + y = 14

Therefore, the equation of line is √3x + y = 14

Question 8: Find the value of θ and p if the equation xcosθ + ysinθ = p id the normal of the line √3x + y + 2 = 0

Solution:

Given equation of line = √3x + y + 2 = 0 

Which can also be written as -√3x – y = 2 

(-√3/2)x + (-1/2)y = 1

This is same as the equation of line i.e. x cos α + y sin α = p

Therefore, cosθ = -√3/2

sinθ = -1/2

p = 1

Hence, θ = 210° = 7π/6 and p =1

Question 9: Find the equation of the straight line which makes a triangle of area 96√3 with the axes and perpendicular from the origin to it makes an angle of 30° with y-axis.

Solution:

Given: Perpendicular from origin makes an angle of 30° with y-axis.

Therefore, it makes 60° with the x-axis.

Also given area of triangle = 96√3

1/2 × 2p × 2p/√3 = 96√3

p² = (96√3 × √3) / 2 = 48 × 3 = 144

p = 12

By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get

x cos60° + y sin60° = 12

(1/2)x + (√3/2) = 12

x + √3y = 24

Therefore, the equation of line is x + √3y = 24

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