📂Geometry

Prove that the Product of the Slopes of Two Perpendicular Lines is Always -1

Theorem

The product of the slopes of two perpendicular lines is always $-1$.

Explanation

This is a fact that can be very useful in many problems. We introduce two methods of proof.

Proof

1

Use Pythagoras’ theorem. See the figure below.

Suppose the slopes of two perpendicular lines are $a$, $a^{\prime}$. Then, considering the right triangle $\triangle OAA^{\prime}$ as shown above, we obtain the following result by Pythagoras’ theorem.

$$ \begin{align*} && {\overline{OA} }^2 + {\overline{OA^{\prime}} }^2 =&\ {\overline{AA^{\prime}} }^2 \\ \implies && (1+a^2) + (1 + {a^{\prime}}^2) =&\ (a-a^{\prime}) ^2 \\ \implies && a^2 + {a^{\prime}}^2 +2 =&\ a^2 + {a^{\prime}}^2 -2aa^{\prime} \\ \implies && 2 =&\ -2aa^{\prime} \\ \implies && aa^{\prime} =&-1 \end{align*} $$

Therefore, the product of the slopes of two perpendicular lines is $-1$.

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2

The slope of a given line is $\tan \theta$. Then, the slopes of two perpendicular lines can be expressed as $\tan \theta$, $\tan \left( \theta +\dfrac{\pi}{2} \right)$ respectively. From this, we obtain the following result.

$$ \begin{align*} \tan \theta \cdot \tan \left( \theta + \frac{\pi}{2} \right) =&\ \frac{\sin \theta}{\cos \theta} \frac{\sin \left( \theta +\dfrac{\pi}{2} \right) }{\cos \left( \theta +\dfrac{\pi}{2} \right) } \\ =&\ \frac{\sin \theta}{\cos \theta} \left( \frac{\cos \theta}{-\sin\theta } \right) \\ =&\ -1 \end{align*} $$

Therefore, the product of the slopes of two perpendicular lines is $-1$.

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