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find the perimeter of the triangle whose vertices are (-3, - 1),(12, 7)…

Question

find the perimeter of the triangle whose vertices are (-3, - 1),(12, 7), and (6, - 1). write the exact answer. do not round.

Explanation:

Step1: Use distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let $A(-3,-1)$, $B(12,7)$ and $C(6,-1)$.

Step2: Calculate distance $AB$

$AB=\sqrt{(12 - (-3))^2+(7 - (-1))^2}=\sqrt{(12 + 3)^2+(7 + 1)^2}=\sqrt{15^2+8^2}=\sqrt{225 + 64}=\sqrt{289}=17$.

Step3: Calculate distance $BC$

$BC=\sqrt{(6 - 12)^2+(-1 - 7)^2}=\sqrt{(-6)^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10$.

Step4: Calculate distance $AC$

$AC=\sqrt{(6 - (-3))^2+(-1 - (-1))^2}=\sqrt{(6 + 3)^2+( - 1+1)^2}=\sqrt{9^2+0^2}=9$.

Step5: Calculate perimeter

The perimeter $P$ of the triangle is $P=AB + BC+AC=17 + 10+9=36$.

Answer:

$36$