How do you write an equation of a line passing through (3, 2), perpendicular to y=5x + 2?

2 Answers
Jun 26, 2018

Find the negative reciprocal of the slope and, using our point, write the equation in slope-intercept form.

Explanation:

The negative reciprocal of the slope of a line is the slope of the line perpendicular to it. If our original slope is 5, then our new slope is -1/5.

Now, let's plug in our value and solve!

2=-1/5(3)+b
2=-3/5+b
2+3/5=b
13/5=b

y=-1/5x+13/5

Jun 26, 2018

I got: y=-1/5x+13/5

Explanation:

Given our line:

y=5x+2

the slope m will be numerical coefficient of x that is: m=5
We know that the slope m' of the perpendiclar to our line must be:

m'=-1/m=-1/5

and the equation of the line with slope m' and passing through our point of coordinates (x_0,y_0) will be:

y-y_0=m'(x-x_0)

in our case:

y-2=-1/5(x-3)

y=-1/5x+3/5+2

y=-1/5x+13/5