Question

Triangle ABC has vertices at A (1,-3), B (3,0)x and C (5,-3). What is the distance from vertex B to the midpoint of AC?

Answer 1

Answer:

The distance from vertex B to the midpoint of AC is 3 units

Step-by-step explanation:

we are given a triangle with vertices  A (1,-3), B (3,0) and C (5,-3).

Mid point of of AC  (x,y) =  ( 1+5/2 , -3 -3/2)

                                     = ( 6/2 , -6/2)

                                     = (3 , -3)

Distance formula = $\sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

                              = $\sqrt{(3-3)^2 + (0 + 3) ^2}$

                               = $\sqrt{(3)^2}$

                                = 3 units