What is the perimeter of a triangle ABC on a graph? A (6,1) B (2,7) C (-3,-5)

Aug 22, 2015

$13 + 5 \sqrt{13}$

Explanation:

Let's see what this triangle looks like.

I used desmos.com to make the graph; it's a great free online graphing calculator!

Anyway, let's use the Pythagorean theorem to find each of the sides. Let's start with the side connecting (-3, -5) and (2, 7). If you go "over" 5 along the x-axis, and "up" 12 along the y-axis, you get from (-3, -5) to (2, 7). So, this side can be thought of as the hypotenuse of a right triangle with legs of 5 and 12.

${5}^{2} + {12}^{2} = {x}^{2}$
$169 = {x}^{2}$
$13 = x$

So this side has length 13. Now let's find the length of the side connecting (2, 7) and (6, 1). To get from (2, 7) to (6, 1), you go "down" 6 and "over" 4. So, this side is the hypotenuse of a right triangle with sides of 6 and 4.

${6}^{2} + {4}^{2} = {x}^{2}$
$52 = {x}^{2}$
$2 \sqrt{13} = x$

So this side has length $2 \sqrt{13}$. One last side (the one from (-3, -5) to (6, 1)). To get from (-3, -5) to (6, 1) you go "over" 9 and "up" 6. So, this side is the hypotenuse of a right triangle with sides of 9 and 6.

${9}^{2} + {6}^{2} = {x}^{2}$
$117 = {x}^{2}$
$3 \sqrt{13} = x$

So this side has length $3 \sqrt{13}$.

This means the total perimeter is 13 + $2 \sqrt{13}$ + $3 \sqrt{13}$ or $13 + 5 \sqrt{13}$.