# Slope of Perpendicular Lines Theorem

If you have already learned about systems of linear equations, then you have probably discussed that the product of the slope of perpendicular lines is $-1$. The proof of this theorem comes from the fact that any point $(x,y)$ rotated 90 degrees at about the origin becomes $(-y,x)$. One example of this is shown below. The point $(3, 4)$, when rotated $90$ degrees counterclockwise becomes $(-4,3)$.

With this fact, we prove this theorem.

Slope of Perpendicular Lines Theorem

If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1 m_2 = -1$

Proof

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be points on line $PQ$ passing through the origin. If we rotate the points on the origin, then, the new coordinates of the points will be $P^\prime= (-y_1, x_1)$ and $Q^\prime(-y_2, x_2)$.

If we let $m_1$ be the slope of $PQ$ and $m_2$ be the slope of $P^\prime Q^\prime$ , then $m_1 = \displaystyle \frac{y_2 - y_1}{x_2 - x_2}$ $m_2 = \displaystyle \frac{x_2 - x_1}{-y_2 - (-y_1)} = \frac{x_2 - x_1}{-(y_2 - y_1)} = -\frac{x_2 - x_1}{y_2 - y_1}$

Multiplying $m_1$ and $m_2$, we have $m_1 m_2 = \left ( \displaystyle \frac{y_2 - y_1}{x_2 - x_1} \right ) \left ( \displaystyle -\frac{x_2 - x_1}{y_2 - y_1} \right )$. $= - 1$

That proves the theorem above.