Answer:
D(4,-3)
Step-by-step explanation:
Given three of the vertices of the square: A(4, -7), B(8, -7),C(8, -3)
Let the coordinate of the fourth vertex be D(x,y).
We know that diagonals of a square are perpendicular bisector. So, the midpoint of both diagonals is the same.
The diagonals are BD and AC
Midpoint of BD = Midpoint of AC
[tex]\left(\dfrac{8+x}{2},\dfrac{-7+y}{2}\right) =\left(\dfrac{4+8}{2},\dfrac{-7+(-3)}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(\dfrac{12}{2},\dfrac{-10}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(6,-5\right)\\$Therefore$:\\\dfrac{8+x}{2}=6\\8+x=12\\x=12-8\\x=4\\$Similarly$\\\dfrac{y-7}{2}=-5\\y-7=-5*2\\y-7=-10\\y=-10+7=-3[/tex]
The coordinates of the fourth vertex is D(4,-3)
