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

In this section, we delve into Chapter 23 of the Class 11 RD Sharma textbook, which focuses on The Straight Lines. Exercise 23.7 is aimed at enhancing students' understanding of the various properties and equations related to straight lines, including slope, intercepts, and different forms of line equations.

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

This section provides comprehensive solutions for Exercise 23.7 from Chapter 23 of the Class 11 RD Sharma textbook. These solutions are designed to help students grasp the core principles of straight lines, ensuring a solid foundation in analytical geometry for further studies.

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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