Question

find the perimeter of the triangle whose vertices are (-10, - 14), (2, 2), and (-10, 7). write the exact answer. do not round. answer 2 points

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 $(-10,-14)$ and $(2,2)$

Let $(x_1,y_1)=(-10,-14)$ and $(x_2,y_2)=(2,2)$. Then $d_1=\sqrt{(2 - (-10))^2+(2-(-14))^2}=\sqrt{(2 + 10)^2+(2 + 14)^2}=\sqrt{12^2+16^2}=\sqrt{144 + 256}=\sqrt{400}=20$.

Step3: Calculate distance between $(-10,-14)$ and $(-10,7)$

Let $(x_1,y_1)=(-10,-14)$ and $(x_2,y_2)=(-10,7)$. Then $d_2=\sqrt{(-10-(-10))^2+(7 - (-14))^2}=\sqrt{0^2+(7 + 14)^2}=\sqrt{21^2}=21$.

Step4: Calculate distance between $(2,2)$ and $(-10,7)$

Let $(x_1,y_1)=(2,2)$ and $(x_2,y_2)=(-10,7)$. Then $d_3=\sqrt{(-10 - 2)^2+(7 - 2)^2}=\sqrt{(-12)^2+5^2}=\sqrt{144 + 25}=\sqrt{169}=13$.

Step5: Calculate perimeter

The perimeter $P$ of a triangle is $P=d_1 + d_2 + d_3$. So $P=20+21 + 13=54$.

Answer:

$54$