**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.

**Solution:**

The equation of the given line is $\frac{x}{4}+\frac{y}{6}=1$.

This equation can also be written as $3 x+2 y-12=0$

$y=\frac{-3}{2} x+6$, which is of the form $y=m x+c$

$\therefore$ Slope of the given line $=-\frac{3}{2}$

Let the given line intersect the $y$-axis at $(0, y)$.

On substituting $x$ with 0 in the equation of the given line, we obtain $\frac{y}{6}=1 \Rightarrow y=6$

$\therefore$ The given line intersects the $y$-axis at $(0,6)$.

The equation of the line that has a slope of $\frac{2}{3}$ and passes through point $(0,6)$ is

$(y-6)=\frac{2}{3}(x-0)$

$3 y-18=2 x$

$2 x-3 y+18=0$

Thus, the required equation of the line is $2 x-3 y+18=0$.