Question

Find the perimeter of triangle whose vertices are
A) (-2, 1)
B) (4, 6)
C) (6, -3)

Answer 1

Plzz see attached image

Answer 2

Perimeter of the given triangle = $\sqrt{61}+\sqrt{85}+4\sqrt{5}$

Solution:

We know the distance between two points is

$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

So the length of AB = $\sqrt{(4-(-2))^{2}+(6-1)^{2}}=\sqrt{(6)^{2}+(5)^{2}}=\sqrt{36+25}=\sqrt{61}$

The length of BC = $\sqrt{(6-4)^{2}+(-3-6)^{2}}=\sqrt{(2)^{2}+(-9)^{2}}=\sqrt{4+81}=\sqrt{85}$

The length of AC = $\sqrt{(6-(-2))^{2}+(-3-1)^{2}}=\sqrt{(8)^{2}+(-4)^{2}}=\sqrt{64+16}=\sqrt{80}=\sqrt{16\times 5}=4 \sqrt{5}$

So, the perimeter of the triangle = $\sqrt{61}+\sqrt{85}+4\sqrt{5}$