Question

find the perimeter of the triangle whose vertices are (-2, -1), (3, 11), and (19, -1). 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 $(x_1,y_1)=(-2,-1)$ and $(x_2,y_2)=(3,11)$. Then $d_1=\sqrt{(3 + 2)^2+(11 + 1)^2}=\sqrt{5^2+12^2}=\sqrt{25 + 144}=\sqrt{169}=13$.

Step3: Calculate the length of the second - side

Let $(x_1,y_1)=(3,11)$ and $(x_2,y_2)=(19,-1)$. Then $d_2=\sqrt{(19 - 3)^2+(-1 - 11)^2}=\sqrt{16^2+(-12)^2}=\sqrt{256+144}=\sqrt{400}=20$.

Step4: Calculate the length of the third - side

Let $(x_1,y_1)=(-2,-1)$ and $(x_2,y_2)=(19,-1)$. Then $d_3=\sqrt{(19 + 2)^2+(-1+1)^2}=\sqrt{21^2+0^2}=21$.

Step5: Calculate the perimeter

The perimeter $P=d_1 + d_2 + d_3$. So $P=13 + 20+21 = 54$.

Answer:

54