How do you find the slope of a line perpendicular to 3x+y=15?

Jun 17, 2015

Subtracting $3 x$ from both sides you get: $y = - 3 x + 15$, which is in standard slope-intercept form, with slope $m = - 3$.

Any line perpendicular to it will have slope $- \frac{1}{m} = \frac{1}{3}$

Explanation:

Either just use the answer above, asserting the $- \frac{1}{m}$ formula, or show it geometrically as follows:

Starting with $3 x + y = 15$, first reflect the line in the ${45}^{o}$ line $x = y$, by swapping $x$ and $y$...

$3 y + x = 15$

Then reflect in the $y$ axis by reversing the sign of $x$

$3 y - x = 15$

These two geometric steps are equivalent to rotating by a right angle centred on the origin.

Next add $x$ to both sides to get:

$3 y = x + 15$

Finally divide both sides by $3$ to get:

$y = \frac{1}{3} x + 5$

This is in slope-intercept format with slope $\frac{1}{3}$