Find the equation of the straight line perpendicular to 2x – 3y = 5

1.	Find the equation of the straight line perpendicular to 2x – 3y = 5 and cutting off an intercept 1 on the positive direction of the x-axis.
Answer» Given: equation is perpendicular to 2x − 3y = 5 and cutting off an intercept 1 on the positive direction of the x-axis. Explanation: The line perpendicular to 2x − 3y = 5 is 3x + 2y + λ = 0 It is given that the line 3x + 2y + λ = 0 cuts off an intercept of 1 on the positive direction of the x axis. This means that the line 3x + 2y + λ = 0 passes through the point (1, 0). ∴ 3 + 0 + λ = 0 ⇒ λ = -3 Substituting the value of λ, we get 3x + 2y – 3 = 0, Hence, equation of the required line.

Find the equation of the straight line perpendicular to 2x – 3y = 5 and cutting off an intercept 1 on the positive direction of the x-axis.

Answer»

Given: equation is perpendicular to 2x − 3y = 5 and cutting off an intercept 1 on the positive direction of the x-axis.

Explanation:

The line perpendicular to 2x − 3y = 5 is

3x + 2y + λ = 0

It is given that the line 3x + 2y + λ = 0 cuts off an intercept of 1 on the positive direction of the x axis.

This means that the line 3x + 2y + λ = 0 passes through the point (1, 0).

∴ 3 + 0 + λ = 0

⇒ λ = -3

Substituting the value of λ, we get 3x + 2y – 3 = 0,

Hence, equation of the required line.