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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

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

1 Answer

  • 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$
answered May 20, 2014 by thanvigandhi_1

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