If triangle ABC has the following verticies, A(-3,7), B(-3,2), C(4,2), what is the length of AC rounded to the nearest tenth?

Answer 1

use the distance formula, the Pythagorean Theorem

AC^2 = (-3-4)^2 + (7-2)^2 = 7^2 + 5^2 = 49+25 = 74

AC = sqr74 = about 8.60 rounded to nearest tenth