Question

what is the length of the longest side of a triangle that has vertices at (-3,-1), (2,-1), and (-3,-7)?

Answer 1

Answer:

$\large\boxed{\sqrt{61}\approx7.81}$

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have the points:

A(-3, -1), B(2, -1) and C(-3, -7).

Substitute:

$AB=\sqrt{(2-(-3))^2+(-1-(-1))^2}=\sqrt{5^2+0^2}=\sqrt{25}=5\\\\AC=\sqrt{(-3-(-3))^2+(-7-(-1))^2}}=\sqrt{0^2+(-6)^2}=\sqrt{36}=6\\\\BC=\sqrt{(-3-2)^2+(-7-(-1))^2}=\sqrt{(-5)^2+(-6)^2}\\=\sqrt{25+36}=\sqrt{61}\\\\\sqrt{61}>6\ \text{because}\ 6^2=36<61$