We are asked to show that the product of the slopes of perpendicular lines is $- 1 .$

Let's use the notation on the diagram.

Point $C$ is the intersection of the two lines.
Points $A$ and $B$ are on the lines so that $\overline{A B}$ is parallel to the $x -$ axis.
Segment $\overline{C D}$ is parallel to the $y -$ axis.

We know that the slope of a line is the quotient of the rise and the run between any two points on the line. Let's use the measurements given on the diagram to write an expression for the slopes.

Points on the Line	Rise	Run	Slope
$A$ and $C$	$b$	$a$	$m_{A C} = \frac{b}{a}$
$C$ and $B$	$- b$	$c$	$m_{C B} = \frac{- b}{c}$

We can use these expressions to find the product of the slopes.

m_{A C} m_{C B} = \frac{b}{a} \cdot \frac{- b}{c} = - \frac{b ^{2}}{a c}

Let's keep this expression in mind. If the lines are perpendicular, then the following statements are also true.

Segment $\overline{A B}$ is the hypotenuse of the right triangle $△ A B C .$
Segment $\overline{C D}$ is the altitude to the hypotenuse in triangle $△ A B C .$

According to Corollary

1

to Theorem

7 - 3

, the length of the altitude is the geometric mean of the lengths of the segments of the hypotenuse.

\frac{a}{b} = \frac{b}{c}

The Cross Products Property tells us that the product of the means is equal to the product of the extremes.

b^{2} = a c

Let's substitute this in the expression for the product of the slopes we got above.

m_{A C} m_{C B} = - \frac{b ^{2}}{a c} = - \frac{a c}{a c} = - 1

This proves that the product of the slopes of perpendicular lines is

- 1 .

Exercise