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

1 Answer
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.