Question

What is the slop of a line perpendicular to the line whose equation is 4x+6y=108. Reduce your answer

Answer 1

The slope of a line perpendicular to the line whose equation is 4x+6y=108 is
`(3)/(2)`

Solution:

Given, line equation is 4x + 6y = 108.

We have to find the slope of the line which is perpendicular to the given line equation.

We know that, product of slopes of perpendicular lines equals to – 1

So, now, let us find the slope of the given line equation.

`\text { Slope of a line }=\frac{-x \text { coefficient }}{y \text { coefficient }}=(-4)/(6)=(-2)/(3)`

Now,

slope of given line
`*` slope of its perpendicular line = -1

`(-2)/(3) * slope of perpendicular line = -1`

`\text { Slope of perpendicular line }=-1 * (3)/(-2)=(-3)/(-2)=(3)/(2)`

Hence, the slope of the perpendicular line is
`(3)/(2)`