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 $latex -1$. The proof of this theorem comes from the fact that any point $latex (x,y)$ rotated 90 degrees at about the origin becomes $latex (-y,x)$. One example of this is shown below. The point $latex (3, 4)$, when rotated $latex 90$ degrees counterclockwise becomes $latex (-4,3)$.

With this fact, we prove this theorem.

Slope of Perpendicular Lines Theorem

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

Proof

Let $latex P(x_1, y_1)$ and $latex Q(x_2, y_2)$ be points on line $latex PQ$ passing through the origin.

If we rotate the points on the origin, then, the new coordinates of the points will be

$latex P^\prime= (-y_1, x_1)$ and $latex Q^\prime(-y_2, x_2)$.

If we let $latex m_1$ be the slope of $latex PQ$ and $latex m_2$ be the slope of $latex P^\prime Q^\prime$ , then

$latex m_1 = \displaystyle \frac{y_2 – y_1}{x_2 – x_2}$

$latex m_2 = \displaystyle \frac{x_2 – x_1}{-y_2 – (-y_1)} = \frac{x_2 – x_1}{-(y_2 – y_1)} = -\frac{x_2 – x_1}{y_2 – y_1}$

Multiplying $latex m_1$ and $latex m_2$, we have

$latex m_1 m_2 = \left ( \displaystyle \frac{y_2 – y_1}{x_2 – x_1} \right ) \left ( \displaystyle -\frac{x_2 – x_1}{y_2 – y_1} \right )$.

$latex = – 1$

That proves the theorem above.