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

Explanation

The line perpendicular to 2x − 3y = 5


is `3x + 2y + lambda = 0`

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


This means that the line `3x + 2y + lambda = 0`

passes through the point `(1, 0)`.

` 3 + 0 + lambda = 0 ⇒ lambda = -3`
Substituting the value of lambda, we
get` 3x + 2y – 3 = 0, `

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