Chapter 1.3, Problem 68HP

To determine

To evaluate the relation between pythagorean theorem and distance formula

Answer to Problem 68HP

Pythagorous theorem is square root of sum of squares of the two distances.

Explanation of Solution

Given information:

Pythagorean theorem and distane formula.

Let, a triangle be $\Delta abc$ .

  

Hence, distance formula is,

  $AB=\sqrt{{(b2-a2)}^2+{(b1-a1)}^2}$

And

  $AC=\sqrt{{(c2-a2)}^2+{(c1-a1)}^2}$

And

  $BC=\sqrt{{(c2-b2)}^2+{(c1-b1)}^2}$

So, pythagorous theorem is

  $AB=\sqrt{B{C}^2+A{C}^2}$

Substituting the values,

  $AB=\sqrt{{(\sqrt{{(c2-b2)}^2+{(c1-b1)}^2})}^2+{(\sqrt{{(c2-a2)}^2+{(c1-a1)}^2})}^2}$

Hence,

  $pythagorous=\sqrt{dis{t}_1^2+dis{t}_2^2}$

Hence, pythagorous theorem is square root of sum of squares of the two distances.