Question

How do you find the equation, in standard form, of the line perpendicular to `2x + 3y = -5` and passing through (3, -5)?

Answer 1

The first information that we have is that we're looking for a line passing through `(3,-5)`. The family of lines passing through a given point `(x_0,y_0)` is `y-y_0=m(x-x_0)`, where `m` is the slope of the line.

So, the lines passing through `(3,-5)` are of the form

`y+5 = m_p(x-3)`, where i called the slope `m_p` for perpendicular.

To find `m_p`, let's find the slope `m` of the original line before: bringing into standard form the equation `2x+3y=-5`, we get `y=\frac{-2x-5}{3} = -2/3 x - 5/3`. So, the original line has slope `-2/3`.

Now, two lines are perpendicular if their slopes `m` and `m'` are in the relation `m=-1/{m'}`.

From this relation, we have `m_p=3/2`, and the solution is thus

`y+5 = 3/2 (x-3)`

Here's a link where you can check that the two lines are perpendicular, and that the second one passes through `(3,5)`.