College Algebra, Chapter 2, 2.1, Section 2.1, Problem 38

Plot the points $A(5,1), B(0,6)$ and $C(-5,1)$ on a coordinate plane. Where must the point $D$ be located so that the quadrilateral $ABCD$ is a square? Find the area of this square.

If the Quadrilateral $ABCD$ is a square, then $d_{AB} = d_{BC} = d_{CD} = d_{AD}$

By using distance formula,

$
\begin{equation}
\begin{aligned}
d_{AB} &= \sqrt{(6-1)^2 + (0-5)^2}\\
\\
&= \sqrt{5^2 + (-5)^2}\\
\\
&= \sqrt{25+25}\\
\\
&= \sqrt{50}\\
\\
&= 5\sqrt{2} \text{ units}\\
\\
d_{AB} &= \sqrt{(y-1)^2 + (x-5)^2} && \text{Distance of point } A(5,1) \text{ and } B(x,y)\\
\\
(d_{AD})^2 &= (y-1)^2 + (x-5)^2 && \text{Square both sides}\\
\\
(5\sqrt{2})^2 &= (y-1)^2 + (x-5)^2 && \text{Substitute } d_{AB} \text{ to } d_{AD}\\
\\
(y-1)^2 &= 50 - (x-5)^2\\
\\
d_{CD} &= \sqrt{(x-(-5))^2 + (y-1)^2} && \text{Distance of point } C(-5,1) \text{ and } D(x,y)\\
\\
(d_{CD})^2 &= (x+5)^2 + (y-1)^2 && \text{Square both sides}\\
\\
(5\sqrt{2})^2 &= (x+5)^2 + (y-1)^2 && \text{Substitute } d_{AB} \text{ to } d_{CD}\\
\\
50 &= (x+5)^2 + \left[50 - (x-5)^2 \right] && \text{Substitute } (y-1)^2 \text{ from } d_{AD}\\
\\
0 &= (x+5)^2 - (x-5)^2 && \text{Subtract to } 50\\
\\
0 &= x^2 + 10x + 25 - x^2 + 10x - 25 && \text{Expand} \\
\\
0 &= 20x && \text{Combine like terms}\\
\\
0 &= x && \text{Solve for } x
\end{aligned}
\end{equation}
$


If $x=0$, then

$
\begin{equation}
\begin{aligned}
(y-1)^2 &= 50-(0-5)^2\\
\\
(y-1)^2 &= 50-25\\
\\
(y-1)^2 &= 25\\
\\
y -1 &= \pm 5\\
\\
y &= \pm 5 + 1\\
\\
y &= 6 \text{ and } y = -4
\end{aligned}
\end{equation}
$

The point $(0,6)$ is already $B$. Therefore, point $D$ is $(0,-4)$
Thus, the area of the square $ABCD$ is $A = \left( d_{AB} \right)^2 = \left( 5\sqrt{2} \right)^2 = 50$ square units.