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find the perimeter of the triangle whose vertices are (-3, - 6), (5,9),…

Question

find the perimeter of the triangle whose vertices are (-3, - 6), (5,9), and (-3,3). 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 side 1 length

Let $(x_1,y_1)=(-3,-6)$ and $(x_2,y_2)=(5,9)$. Then $d_1=\sqrt{(5 - (-3))^2+(9 - (-6))^2}=\sqrt{(8)^2+(15)^2}=\sqrt{64 + 225}=\sqrt{289}=17$.

Step3: Calculate side 2 length

Let $(x_1,y_1)=(5,9)$ and $(x_2,y_2)=(-3,3)$. Then $d_2=\sqrt{(-3 - 5)^2+(3 - 9)^2}=\sqrt{(-8)^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=10$.

Step4: Calculate side 3 length

Let $(x_1,y_1)=(-3,3)$ and $(x_2,y_2)=(-3,-6)$. Then $d_3=\sqrt{(-3-(-3))^2+(-6 - 3)^2}=\sqrt{(0)^2+(-9)^2}=\sqrt{81}=9$.

Step5: Calculate perimeter

The perimeter $P$ of a triangle is $P=d_1 + d_2 + d_3$. So $P=17+10 + 9=36$.

Answer:

36