Question

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

Answer 1

Answer:

The slope of a line perpendicular to the line whose equation is 4x+6y=108 is $\frac{3}{2}$

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.

$\text { Slope of a line }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{6}=\frac{-2}{3}$

Now,

slope of given line $\times$ slope of its perpendicular line = -1

$\frac{-2}{3} \times slope of perpendicular line = -1$

$\text { Slope of perpendicular line }=-1 \times \frac{3}{-2}=\frac{-3}{-2}=\frac{3}{2}$

Hence, the slope of the perpendicular line is $\frac{3}{2}$