Question

What is the length of the longest side of a triangle that has the vertices (-5, 6), (-5, -2), and (1, -2)? Round to the nearest hundredth, if necessary

Answer 1

Well you'd use the distance formula for each pair of points:

d = sq.root of((x2 - x1)^2 + (y2-y1)^2)

(1) For the points (-5,6) and (-5,-2):

d = sq.root of((-5-(-5))^2 + (-2-6)^2)
= sq.root of (0 + 64)
= 8

(2) For the points (-5,-2) and (1,-2):

d = sq.root of((1-(-5))^2 + (-2-(-2))^2)
= sq.root of (36 + 0)
= 6

(3) For the points (1,-2) and (-5,6):

d = sq.root of((-5-1)^2 + (6-(-2))^2)
= sq.root of(36 + 64)
= sq.root of(100)
= 10

Therefor the length of the longest side is 10 units