Question

find the perimeter of the triangle whose vertices are (1, - 3), (-5,0), and (-3, - 8). write the exact answer. do not round.

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 distance between $(1,-3)$ and $(-5,0)$

Let $(x_1,y_1)=(1,-3)$ and $(x_2,y_2)=(-5,0)$. Then $d_1=\sqrt{(-5 - 1)^2+(0+ 3)^2}=\sqrt{(-6)^2+3^2}=\sqrt{36 + 9}=\sqrt{45}=3\sqrt{5}$.

Step3: Calculate distance between $(-5,0)$ and $(-3,-8)$

Let $(x_1,y_1)=(-5,0)$ and $(x_2,y_2)=(-3,-8)$. Then $d_2=\sqrt{(-3 + 5)^2+(-8 - 0)^2}=\sqrt{2^2+(-8)^2}=\sqrt{4 + 64}=\sqrt{68}=2\sqrt{17}$.

Step4: Calculate distance between $(-3,-8)$ and $(1,-3)$

Let $(x_1,y_1)=(-3,-8)$ and $(x_2,y_2)=(1,-3)$. Then $d_3=\sqrt{(1 + 3)^2+(-3 + 8)^2}=\sqrt{4^2+5^2}=\sqrt{16 + 25}=\sqrt{41}$.

Step5: Calculate perimeter

The perimeter $P$ of the triangle is $P=d_1 + d_2 + d_3=3\sqrt{5}+2\sqrt{17}+\sqrt{41}$.

Answer:

$3\sqrt{5}+2\sqrt{17}+\sqrt{41}$