Chapter 15.2, Problem 23PSC

To determine

To Prove: The shortest segment between a point and the line is the segment perpendicular to the line.

Explanation of Solution

Given:

  $\begin{array}{l}\overrightarrow{AP}\text{ is the perpendicular to }\overleftrightarrow{m}\end{array}$

Formula Used:

Pythagorean Theorem:

  $\begin{array}{l}\text{For right angle triangles}\\ {\text{a}}^2+b^2=c^2\end{array}$

Calculation:

  $\begin{array}{l}\text{Let us assume that }\overrightarrow{AP}\text{ is not the shortest distance}\text{.}\\ \text{⇒}\text{there is a shorter segment }\overrightarrow{PB}\text{ which is not a right angle to }\overleftrightarrow{m}\\ \Delta PAB\text{ is a right angle triangel and }\overrightarrow{PB}\text{ is the hypotenuse}\text{.}\\ \text{⇒}\text{Hypotenuse can't be smaller than the side,}\\ \text{⇒}\overrightarrow{AP}<\overrightarrow{PB}\\ \text{ and it is against our assumption }\\ \text{⇒}\overrightarrow{AP}\text{ is the shortest distance}\text{.}\end{array}$

Conclusion:

Hence Proved.