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# Find the equation of a line drawn perpendicular to the line x/4+y/6=1 through the point, where it meets the y-axis.

Q : 7      Find the equation of a line drawn perpendicular to the line  $\small \frac{x}{4}+\frac{y}{6}=1$
through the point, where it meets the $\small y$-axis.

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given equation of line is
$\small \frac{x}{4}+\frac{y}{6}=1$
we can rewrite it as
$3x+2y=12$
Slope of line $3x+2y=12$ , $m' = -\frac{3}{2}$
Let the Slope of perpendicular line is m
$m = -\frac{1}{m'}= \frac{2}{3}$
Now, the ponit of intersection of $3x+2y=12$  and $x =0$  is   $(0,6)$
Equation of line passing through point $(0,6)$ and with slope $\frac{2}{3}$  is
$(y-6)= \frac{2}{3}(x-0)$
$3(y-6)= 2x$
$2x-3y+18=0$
Therefore, equation of line is   $2x-3y+18=0$

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