Question

What is the slope of a line that is perpendicular to the line whose equation is 8y−5x=118y−5x=11?

Answer 1

8y−5x=11 
8y = 5x + 11
y = 5/8x + 11/8 has slope = 5/8
a line that is perpendicular to the line, slope is opposite and reciprocal so slope = - 8/5

answer
slope = -8/5

Answer 2

Answer:

The slope of a line that is perpendicular to the line whose equation is  $8y-5x=11$ is $m=-\frac{8}{5}$                      

Step-by-step explanation:

Given : Equation $8y-5x=11$

To find : What is the slope of a line that is perpendicular to the line whose equation is given?

Solution :

First we find the slope of the given line,

The general slope form of line is $y=mx+b$ where m is the slope of the line and b is the y-intercept.

Re-write the given equation into general form,

$8y-5x=11$

Take 5x to another side,

$8y=5x+11$

Divide both side by 8,

$y=\frac{5}{8}x+\frac{11}{8}$

On comparing with general form,

The slope of the line is $m=\frac{5}{8}$

We know,

When two line are perpendicular one slope is negative reciprocal of another.

If the slope of line is $m=\frac{5}{8}$

Then the slope of perpendicular line on this line is  $m=-\frac{8}{5}$

Therefore, The slope of a line that is perpendicular to the line whose equation is  $8y-5x=11$ is $m=-\frac{8}{5}$