Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point

It is given that

\( \dfrac{x}{4}+\dfrac{y}{6}=1\)

We can write it as

3x + 2y – 12 = 0

So we get

y = \(- \dfrac{3}{2}x\)+ 6, which is of the form y = mx + c

Here the slope of the given line = \( - \dfrac{3}{2}x\)

So the slope of line perpendicular to the given line 

=  =\(\dfrac{-1}{ - \dfrac{3}{2}}= \dfrac{2}{3} \)

Consider the given line intersect the y-axis at (0, y)

By substituting x as zero in the equation of the given line

\( \dfrac{y}{6}\) = 1

y = 6

Hence, the given line intersects the y-axis at (0, 6)

We know that the equation of the line that has a slope of\( \dfrac{2}{3}\) and passes through point (0, 6) is

(y – 6) = \( \dfrac{2}{3}\)(x – 0)

By further calculation

3y – 18 = 2x

So we get

2x – 3y + 18 = 0

Hence, the required equation of the line is 2x – 3y + 18 = 0.