Question

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 1

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).

$$\begin{gathered} \therefore 3+0+\lambda =0 \\ \Rightarrow \lambda =-3 \end{gathered}$$

Substituting the value of $\lambda$, we get $3x+2y-3=0$, which is equation of the required line.