The shortest distance theorem states that the shortest distance between a point P, and a line, l, is the perpendicular line from P to l. It is also called the “perpendicular distance.”

It is simple to prove this theorem using the Pythagorean Theorem.

## Problem

Prove that the shortest distance between a point P, and a line, l, is the perpendicular line from P to l.

## Strategy

Since the perpendicular line from P to l forms a right angle, we will try to use what we know about right triangles, and the theorem we have about lengths of the sides of right triangle – the Pythagorean Theorem.

Any line from P to l other than the perpendicular line PA will form a right triangle, where that new line is the hypotenuse.

From the Pythagorean Theorem, we know that PB^{2}= PA^{2}+ AB^{2}. Since both PA and AB have non-zero length, their squares are positive numbers.

If PB^{2} is equal to PA^{2} plus some positive number (AB^{2}), it is bigger than PA^{2} and thus PB > PA, for every point B other than the one where the perpendicular line meets the line l.

## Proof

(1) PA ⊥l //Given

(2) ΔPAB is a right triangle //Definition of a right triangle, from (1)

(3) PB^{2}= PA^{2}+ AB^{2} //Pythagorean Theorem

(4) AB^{2}>0 //Square of a non-zero number is positive

(5) PB^{2}>PA^{2} //(3),(4)

(6) PB>PA //Lines have positive lengths, so the square root is positive.