Question

You are graphing Square ABCDABCDA, B, C, D in the coordinate plane. The following are three of the vertices of the square: A(4, -7), B(8, -7),A(4,−7),B(8,−7),A, left parenthesis, 4, comma, minus, 7, right parenthesis, comma, B, left parenthesis, 8, comma, minus, 7, right parenthesis, comma and C(8, -3)C(8,−3)C, left parenthesis, 8, comma, minus, 3, right parenthesis. What are the coordinates of point DDD? \large((left parenthesis , \large))right parenthesis

Answer 1

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

$\left(\dfrac{8+x}{2},\dfrac{-7+y}{2}\right) =\left(\dfrac{4+8}{2},\dfrac{-7+(-3)}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(\dfrac{12}{2},\dfrac{-10}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(6,-5\right)\\ Therefore :\\\dfrac{8+x}{2}=6\\8+x=12\\x=12-8\\x=4\\ Similarly \\\dfrac{y-7}{2}=-5\\y-7=-5*2\\y-7=-10\\y=-10+7=-3$

The coordinates of the fourth vertex is D(4,-3)