Question

find the perimeter of the triangle whose vertices are (1, - 4), (-7,0), and (-4, - 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 distance between $(1,-4)$ and $(-7,0)$

Let $(x_1,y_1)=(1,-4)$ and $(x_2,y_2)=(-7,0)$. Then $d_1=\sqrt{(-7 - 1)^2+(0+4)^2}=\sqrt{(-8)^2 + 4^2}=\sqrt{64 + 16}=\sqrt{80}=4\sqrt{5}$.

Step3: Calculate distance between $(-7,0)$ and $(-4,-3)$

Let $(x_1,y_1)=(-7,0)$ and $(x_2,y_2)=(-4,-3)$. Then $d_2=\sqrt{(-4 + 7)^2+(-3 - 0)^2}=\sqrt{3^2+(-3)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}$.

Step4: Calculate distance between $(-4,-3)$ and $(1,-4)$

Let $(x_1,y_1)=(-4,-3)$ and $(x_2,y_2)=(1,-4)$. Then $d_3=\sqrt{(1 + 4)^2+(-4 + 3)^2}=\sqrt{5^2+(-1)^2}=\sqrt{25 + 1}=\sqrt{26}$.

Step5: Calculate perimeter

The perimeter $P=d_1 + d_2 + d_3=4\sqrt{5}+3\sqrt{2}+\sqrt{26}$.

Answer:

$4\sqrt{5}+3\sqrt{2}+\sqrt{26}$