Vectors – How to Calculate Distance of a Point from a Straight Line

analytic geometryeducationvectors

For the straight line point distance there are several proofs. Several years ago I have proved (but I no longer remember the proof), the point-right distance by remembering that for a line $$r: ax+by+c=0$$ $\mathbf{ \hat n}=(a,b)$ is perpendicular to $r$ .I kindly ask if someone can answer me with simple proof for 15-year-old students in high school of a scientific high school. Thank you very much.

Best Answer

Consider the picture below.

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On one hand the area of the triangle $ABC$ is $$\frac{1}{2}|BC||AC|=\frac{1}{2}|y_1-y_0||x_1-x_0|.$$

On the other hand the same area is $$\frac{1}{2}|AB|h=\frac{1}{2}\sqrt{|y_1-y_0|^2+|x_1-x_0|^2}h,$$ where $h$ is the distance of $C=(x_0,y_0)$ to the line. Hence \begin{equation}|y_1-y_0||x_1-x_0|=\sqrt{|y_1-y_0|^2+|x_1-x_0|^2}h\ \ \ \ (*)\end{equation}

Now, since $A,B$ belong to the line and its equation is $ax+by+c=0$, we have

$$ax_0+by_1+c=0\Rightarrow y_1=\frac{-c-ax_0}{b}$$ $$ax_1+by_0+c=0\Rightarrow x_1=\frac{-c-by_0}{a}$$

Thus, $|y_1-y_0|=\frac{d}{|b|}$ and $|x_1-x_0|=\frac{d}{|a|}$, where $d=|ax_0+by_0+c|$.

Finally, by $(*)$, we have $$\frac{d}{|b|}\frac{d}{|a|}=\sqrt{\frac{d^2}{|b|^2}+\frac{d^2}{|a|^2}}h\ \ \Rightarrow\ \ h=\frac{d}{\sqrt{a^2+b^2}}.$$

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