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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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