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.
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 $2 x-3 y+18=0$.
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