Question

1. a triangle in the coordinate plane has vertices d(-5, 8), e(-2, 3), and f(-5, 3). find the perimeter.

Explanation:

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

For side $DE$: Let $(x_1,y_1)=(-5,8)$ and $(x_2,y_2)=(-2,3)$. Then $d_{DE}=\sqrt{(-2 + 5)^2+(3 - 8)^2}=\sqrt{3^2+( - 5)^2}=\sqrt{9 + 25}=\sqrt{34}$.

Step2: For side $EF$: Let $(x_1,y_1)=(-2,3)$ and $(x_2,y_2)=(-5,3)$. Then $d_{EF}=\sqrt{(-5 + 2)^2+(3 - 3)^2}=\sqrt{(-3)^2+0^2}=3$.

Step3: For side $DF$: Let $(x_1,y_1)=(-5,8)$ and $(x_2,y_2)=(-5,3)$. Then $d_{DF}=\sqrt{(-5+5)^2+(3 - 8)^2}=\sqrt{0^2+( - 5)^2}=5$.

Step4: Calculate the perimeter $P$.

$P=d_{DE}+d_{EF}+d_{DF}=\sqrt{34}+3 + 5=\sqrt{34}+8$.

Answer:

$\sqrt{34}+8$