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

`\large\boxed{√(61)\approx7.81}`

Explanation:

The formula of a distance between two points:

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