Problema Solution
find the lengths of the sides of triangle ABC if the vertices are A(-1,5), B(6,1), C(-2,-1)
Answer provided by our tutors
we will use the distance formula:
the distance d between two points (x1, y1) and (x2, y2) is equal to:
d^2 = (x2 - x1)^2 + (y2 - y1)^2
thus we have
AB^2 = (6 - (-1))^2 + (1 - 5)^2
AB^2 = 49 + 16
AB^2 = 65
AB = 8.06
BC^2 = (-2-6)^2 + (- 1 - 1)^2
BC^2 = 64 + 4
BC^2 = 68
BC = 8.25
AC^2 = (-2 - (-1))^2 + (-1 - 5)^2
AC^2 = 1 + 36
AC^2 = 37
AC = 6.08