Question

a triangle in the coordinate plane has vertices r(-2, 2), s(1, -1), and t(-2, -1). find the perimeter.

Explanation:

Step1: Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ to find the length of $RS$.

$x_1=-2,y_1 = 2,x_2=1,y_2=-1$
$RS=\sqrt{(1-(-2))^2+(-1 - 2)^2}=\sqrt{(3)^2+(-3)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}$

Step2: Find the length of $ST$.

$x_1=1,y_1=-1,x_2=-2,y_2=-1$
$ST=\sqrt{(-2 - 1)^2+(-1-(-1))^2}=\sqrt{(-3)^2+0^2}=3$

Step3: Find the length of $TR$.

$x_1=-2,y_1=-1,x_2=-2,y_2=2$
$TR=\sqrt{(-2-(-2))^2+(2 - (-1))^2}=\sqrt{0^2+(3)^2}=3$

Step4: Calculate the perimeter $P$.

$P=RS + ST+TR=3\sqrt{2}+3 + 3=6 + 3\sqrt{2}$

Answer:

$6 + 3\sqrt{2}$