Find the equation of a line for which(i) p = 5, α = 60°(ii) p = 4, �

1.	Find the equation of a line for which(i) p = 5, α = 60°(ii) p = 4, α = 150°
Answer» (i) p = 5, α = 60° Given: p = 5, α = 60° The equation of the line in normal form is given by Using the formula, x cos α + y sin α = p Now, substitute the values, we get x cos 60° + y sin 60° = 5 x/2 + √3y/2 = 5 x + √3y = 10 ∴ The equation of line in normal form is x + √3y = 10. (ii) p = 4, α = 150° Given: p = 4, α = 150° The equation of the line in normal form is given by Using the formula, x cos α + y sin α = p Now, substitute the values, we get x cos 150° + y sin 150° = 4 cos (180° – θ) = – cos θ , sin (180° – θ) = sin θ x cos(180° – 30°) + y sin(180° – 30°) = 4 – x cos 30° + y sin 30° = 4 –√3x/2 + y/2 = 4 -√3x + y = 8 ∴ The equation of line in normal form is -√3x + y = 8.

Find the equation of a line for which(i) p = 5, α = 60°(ii) p = 4, α = 150°

Answer»

(i) p = 5, α = 60°

Given:

p = 5, α = 60°

The equation of the line in normal form is given by

Using the formula,
x cos α + y sin α = p

Now, substitute the values, we get
x cos 60° + y sin 60° = 5

x/2 + √3y/2 = 5
x + √3y = 10

∴ The equation of line in normal form is x + √3y = 10.

(ii) p = 4, α = 150°

Given:

p = 4, α = 150°

The equation of the line in normal form is given by

Using the formula,
x cos α + y sin α = p

Now, substitute the values, we get

x cos 150° + y sin 150° = 4

cos (180° – θ) = – cos θ , sin (180° – θ) = sin θ

x cos(180° – 30°) + y sin(180° – 30°) = 4

– x cos 30° + y sin 30° = 4

–√3x/2 + y/2 = 4

-√3x + y = 8

∴ The equation of line in normal form is -√3x + y = 8.