Question

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

Explanation:

Step1: Use distance formula for side RS

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For $R(-2,2)$ and $S(1,-1)$, we have $x_1=-2,y_1 = 2,x_2=1,y_2=-1$. Then $RS=\sqrt{(1-(-2))^2+(-1 - 2)^2}=\sqrt{(3)^2+(-3)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}$.

Step2: Use distance formula for side ST

For $S(1,-1)$ and $T(-2,-1)$, with $x_1 = 1,y_1=-1,x_2=-2,y_2=-1$, then $ST=\sqrt{(-2 - 1)^2+(-1-(-1))^2}=\sqrt{(-3)^2+0^2}=3$.

Step3: Use distance formula for side RT

For $R(-2,2)$ and $T(-2,-1)$, with $x_1=-2,y_1 = 2,x_2=-2,y_2=-1$, then $RT=\sqrt{(-2-(-2))^2+(-1 - 2)^2}=\sqrt{0^2+(-3)^2}=3$.

Step4: Calculate perimeter

The perimeter $P$ of a triangle is $P=RS + ST+RT$. Substituting the values, we get $P=3\sqrt{2}+3 + 3=6 + 3\sqrt{2}$.

Answer:

$6 + 3\sqrt{2}$