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

The equation of the given line is $\frac{x}{4}+\frac{y}{6}=1$.
This equation can also be written as $3x+2y-12=0$
$\Rightarrow y=\frac{-3}{2}x+6,$ which is of the form $y=mx+c$
$\therefore$ slope of given line $=-\frac{3}{2}$
$\therefore$ slope of line perpendicular to given line $=-\frac{1}{\{-\frac{3}{2}\}}=\frac{2}{3}$
On substituting $x=0$ in the given equation of line,
we obtain $\frac{y}{6}=1\Rightarrow y=6$
$\therefore$ the given line intersect the $y$-axis at $(0,6)$
Hence, the equation of line that has slope $\frac{2}{3}$ and passes through point $(0,6)$ is
$(y-6)=\frac{2}{3}(x-0)$
$\Rightarrow 3y-18=2x\Rightarrow 2x-3y+18=0$
Thus, the required equation of the line is $2x-3y+18=0$.