Question

Find the equation of the straight line which has y-intercept equal to $\frac{4}{3}$ and is perpendicular to 3x − 4y + 11 = 0.

Answer 1

The line perpendicular to 3x − 4y + 11 = 0 is $4x+3y+\lambda =0$

It is given that the line $4x+3y+\lambda =0$ has y-intercept equal to $\frac{4}{3}$
This means that the line passes through $\left(0,\frac{4}{3}\right)$

$$\begin{gathered} \therefore 0+4+\lambda =0 \\ \Rightarrow \lambda =-4 \end{gathered}$$

Substituting the value of $\lambda$, we get $4x+3y-4=0$, which is equation of the required line.