Question

problem 3: perimeter of a triangle
find the perimeter of the triangle with vertices a(-2, 1), b(-2, 5), and c(4, -1). show calculations/how you found each side length.
work space:

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate length of side AB

For points $A(-2,1)$ and $B(-2,5)$, $x_1=-2,y_1 = 1,x_2=-2,y_2 = 5$.
$AB=\sqrt{(-2+2)^2+(5 - 1)^2}=\sqrt{0 + 16}=4$.

Step3: Calculate length of side BC

For points $B(-2,5)$ and $C(4,-1)$, $x_1=-2,y_1 = 5,x_2=4,y_2=-1$.
$BC=\sqrt{(4 + 2)^2+(-1 - 5)^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}$.

Step4: Calculate length of side AC

For points $A(-2,1)$ and $C(4,-1)$, $x_1=-2,y_1 = 1,x_2=4,y_2=-1$.
$AC=\sqrt{(4 + 2)^2+(-1 - 1)^2}=\sqrt{36 + 4}=\sqrt{40}=2\sqrt{10}$.

Step5: Calculate perimeter

The perimeter $P=AB + BC+AC=4+6\sqrt{2}+2\sqrt{10}$.

Answer:

$4 + 6\sqrt{2}+2\sqrt{10}$