Question

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

Answer 1

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`.