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.

Explanation

Given line , `(x)/(4)+(y)/(6)=1`
`implies 3x+2y=12`……`(1)`
At `y`-axis
`x=0`
`:. 0+2y=12impliesy=6`
This line meets `y`-axis at `(0,6)`
Let the equation of a line perpendicular to line `(1)` is,
`2x-3y=lambda`……`(2)`
This line passes through the point `(0,6)`
`:. 0-18=lambdaimplieslambda=-18`
Form equation `(2)`, equation of line,
`:. 2x-3y=-18implies2x-3y+18=0`