Question

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

• A. $2x-3y+18=0$
• B. $2x+3y-18=0$
• C. $2x-3y-18=0$
• D. None of these

Answer 1

Answer: (A)

Given, equation of line is
$\frac{x}{4}+\frac{y}{6}=1$
$\Rightarrow \frac{3x+2y}{12}=1$
$\Rightarrow 3x+2y=\mathrm{12....}$(i)
If line (i) meets the $Y$-axis, putting $x=0$ in Eq. (i), we get
$\Rightarrow 0+2y=12$
$\Rightarrow y=6$
$\therefore$ Point is $(0,6)$.
Now, equation of line perpendicular to Eq. (i) is
$2x-3y+k=\mathrm{0....}$(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
$2x-3y+18=0$