Question

Find the equation of a line drawn perpendicular to the line $\frac{x}{4}+\frac{y}{6}=1$ through the point where it meets the y-axis.

Answer 1

Let us find the intersection of the line $\frac{x}{4}+\frac{y}{6}=1$ with y-axis.

At x = 0,
$$\begin{gathered} 0+\frac{y}{6}=1 \\ \Rightarrow y=6 \end{gathered}$$

Thus, the given line intersects y-axis at (0, 6).

The line perpendicular to the line $\frac{x}{4}+\frac{y}{6}=1$ is

$\frac{x}{6}-\frac{y}{4}+\lambda =0$

This line passes through (0, 6).

$$\begin{gathered} 0-\frac{6}{4}+\lambda =0 \\ \Rightarrow \lambda =\frac{3}{2} \end{gathered}$$

Now, substituting the value of $\lambda$, we get:
$$\begin{gathered} \frac{x}{6}-\frac{y}{4}+\frac{3}{2}=0 \\ \Rightarrow 2x-3y+18=0 \end{gathered}$$

This is the equation of the required line.