find the slope of a line perpendicular to 6y+13x=22 ?

Question

find the slope of a line perpendicular to the graph of this equation

Answer 1

Given line 6y+13x = 22

Subtract 13x from each side.

6y+13x-13x = 22-13x

6y = -13x+22

Divide to each side by 6.

6y/6 = (-13/6)x+22/6

y = (-13/6)x+22/6

Slope of the given line say m1  = -13/6

Perpendicular line to given line slope say m2.

We know that m1*m2 = -1

So m2 = 6/13

Required slope of the line is 6/13.

Answer 2

The line equation is 6y + 13x = 22.

Write the equation in slope-intercept form line equation is y = mx + b, where m is slope and b is y-intercept.

6y = - 13x + 22

y = (- 13/6)x + (22/6)

y = (- 13/6)x + (11/3).

Compare the equation with slope-intercept form line equation is y = mx + b, where m is slope and b is y-intercept.

Slope (m) = - 13/6.

Because the slopes of perpendicular lines are negative reciprocals, the slope of perpendicular line is 6/13.