How do you write an equation of a line passing through (3, -5), perpendicular to #5x+2y=10#?

1 Answer
Burglar
Jul 4, 2016

It is #y=2/5x-31/5#.

Explanation:

First of all we put the line #5x+2y=10# in the form #y=mx+q#.

#5x+2y=10#

#2y=-5x+10#

#y=-5/2x+5#.

Now it is easy to find all the lines perpendicular to this. It is enough to take the opposite inverse of the slope #m->-1/m#. In this case #m=-5/2# then the slope of the perpendicular lines is #2/5# and the equation is

#y=2/5x+q#.

To find #q# we impose the passage for the point #(3, -5)#

#-5=2/5*3+q#

#-5=6/5+q#

#-5-6/5=q#

#-31/5=q#.

The equation of the perpendicular is finally

#y=2/5x-31/5#.

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