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

2 Answers
Jun 26, 2018

Answer:

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

Answer:

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#