How do you write an equation of a line passing through (3, -5), perpendicular to 5x+2y=105x+2y=10?

1 Answer
Jul 4, 2016

It is y=2/5x-31/5y=25x315.

Explanation:

First of all we put the line 5x+2y=105x+2y=10 in the form y=mx+qy=mx+q.

5x+2y=105x+2y=10

2y=-5x+102y=5x+10

y=-5/2x+5y=52x+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/mm1m. In this case m=-5/2m=52 then the slope of the perpendicular lines is 2/525 and the equation is

y=2/5x+qy=25x+q.

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

-5=2/5*3+q5=253+q

-5=6/5+q5=65+q

-5-6/5=q565=q

-31/5=q315=q.

The equation of the perpendicular is finally

y=2/5x-31/5y=25x315.