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Q.
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, is
Straight Lines
Solution:
Given, equation of line is
$ \frac{x}{4}+\frac{y}{6}=1 $
$\Rightarrow \frac{3 x+2 y}{12}=1 $
$\Rightarrow 3 x+2 y=12 ....$(i)
If line (i) meets the $Y$-axis, putting $x=0$ in Eq. (i), we get
$\Rightarrow 0+2 y =12 $
$\Rightarrow y =6$
$\therefore$ Point is $(0,6)$.
Now, equation of line perpendicular to Eq. (i) is
$2 x-3 y+k=0 ....$(ii)
Line (ii) passes through the point $(0,6)$, so it will satisfy it.
i.e., $ 2 \times 0-3 \times 6+k=0$
$\Rightarrow k=18 (\because$ put $x=0, y=6)$
Hence, Eq. (ii) becomes
$2 x-3 y+18=0$