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

May 25, 2015

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

$x + 3 y = 6$

with

$y - 3 x = 6$

to get a perpendicular line.

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

$y = 3 x + 6$

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 $\left(- 3 , 5\right)$, we can write the equation of the desired line in standard point slope form as:

$y - 5 = 3 \left(x - \left(- 3\right)\right) = 3 \left(x + 3\right)$

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

$y = 3 \left(x + 3\right) + 5 = 3 x + 9 + 5 = 3 x + 14$