Browse Questions

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

$\begin {array} {1 1} (A)\;2x-3y+18=0 & \quad (B)\;3y-2x+18=0 \\ (C)\;2x-3y-18=0 & \quad (D)\;3y-2x-18=0 \end {array}$

Toolbox:
• If two lines are perpendicular, then the product of their slopes is -1. i.e., $m_1m_2=-1$
• Equation of the line with slope $m$ and passing through $(x_1, y_1)$ is $y-y_1=m(x-x_1)$
Given equation of the line is $\large\frac{x}{4}+\large\frac{y}{6}$$=1 This can be written as 3x+2y-12=0 This can be written in the form of y=mx+c (i.e.,) y= \bigg( \large\frac{-3}{2} \bigg)$$x+6$
Hence the slope of the given line $= \large\frac{-3}{2}$
Slope of line perpendicular of the given line
$= \large\frac{-1}{-\bigg(\Large\frac{3}{2}\bigg)}$$= \large\frac{2}{3} It is given that the line intersect the y - axis. Let this point be (0, y) On substituting x with 0 in the equation of the given line, we obtain \large\frac{y}{6}$$=1 \Rightarrow y=6$
Now the equation of the line whose slope is $\large\frac{2}{3}$ and point (0,6) is
$y-6=\large\frac{2}{3} (x-0)$
$\Rightarrow 3y-18=2x$
$\Rightarrow 2x-3y+18=0$
Hence the required equation is $2x-3y+18=0$