Question

How do you write an equation in standard form for a line perpendicular to x+3y=6 and passing through (-3,5)?

Answer 1

Fun fact:

Given an equation in `x` and `y` of a line, then you can construct the equation of a perpendicular line by swapping `x` and `y` and reversing the sign of one of them.

This is equivalent to reflecting the original line in the `45^o` line with equation `y=x` then reflecting in one of the axes.

So in your example, we can replace the original

`x+3y=6`

with

`y-3x=6`

to get a perpendicular line.

Then adding `3x` to both sides we get:

`y=3x+6`

which is in standard slope intercept form, with slope `3` and intercept `6`

The line we want to construct is parallel to this, so will have the same slope, `3`, but probably a different intercept.

Given the slope `3` and the point `(-3, 5)`, we can write the equation of the desired line in standard point slope form as:

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

To convert this to slope intercept form, add `5` to both sides to get:

`y = 3(x+3)+5 = 3x+9+5 = 3x+14`