Question

Find the perimeter of ABC with vertices A (1,1), B (7,1), and C (1,9)

Answer 1

Answer:

24 unit

Step-by-step explanation:

Given,

The vertices of the triangle ABC are,

A (1,1), B (7,1), and C (1,9),

By the distance formula,

$AB=\sqrt{(7-1)^2+(1-1)^2}=\sqrt{6^2}=6\text{ unit}$

$BC=\sqrt{(1-7)^2+(9-1)^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{ unit}$

$CA=\sqrt{(1-1)^2+(1-9)^2}=\sqrt{8^2}=8\text{ unit}$

Thus, the perimeter of the triangle ABC = AB + BC + CA = 6 + 10 + 8 = 24 unit