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question 10 of 13, step 1 of 1 find the perimeter of the triangle whose…

Question

question 10 of 13, step 1 of 1
find the perimeter of the triangle whose vertices are (-8, -2), (1, 10), and (17, -2). write the exact answer. do not round.

Explanation:

Step1: Recall the 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 the length of the first - side

Let $(x_1,y_1)=(-8,-2)$ and $(x_2,y_2)=(1,10)$. Then $d_1=\sqrt{(1-(-8))^2+(10 - (-2))^2}=\sqrt{(1 + 8)^2+(10 + 2)^2}=\sqrt{9^2+12^2}=\sqrt{81 + 144}=\sqrt{225}=15$.

Step3: Calculate the length of the second - side

Let $(x_1,y_1)=(1,10)$ and $(x_2,y_2)=(17,-2)$. Then $d_2=\sqrt{(17 - 1)^2+(-2 - 10)^2}=\sqrt{16^2+(-12)^2}=\sqrt{256+144}=\sqrt{400}=20$.

Step4: Calculate the length of the third - side

Let $(x_1,y_1)=(-8,-2)$ and $(x_2,y_2)=(17,-2)$. Then $d_3=\sqrt{(17-(-8))^2+(-2-(-2))^2}=\sqrt{(17 + 8)^2+( - 2+2)^2}=\sqrt{25^2+0^2}=25$.

Step5: Calculate the perimeter

The perimeter $P$ of a triangle is $P=d_1 + d_2 + d_3$. So $P=15 + 20+25=60$.

Answer:

60