Question

Show that the points $A\left(5,6\right),B\left(1,5\right),C\left(2,1\right),\mathrm{and}D\left(6,2\right)$) are the vertices of a square.

Answer 1

Step 1. Determine whether all the sides of the quadrilateral are equal or not.

Draw the diagram as follows:

Determine the distance between the given vertices to prove the given statement.

The distance formula is given by $\mathbf{D}\mathbf{=}\sqrt{{\left(x_2-x_1\right)}^{\mathbf{2}}\mathbf{+}{\left(y_2-y_1\right)}^{\mathbf{2}}}$.

Apply the distance formula to find the distance between the vertices $A\left(5,6\right)\mathrm{and}B\left(1,5\right)$.

$$\begin{gathered} AB=\sqrt{{\left(1-5\right)}^2+{\left(5-6\right)}^2} \\ AB=\sqrt{{\left(-4\right)}^2+{\left(-1\right)}^2} \\ AB=\sqrt{17} \end{gathered}$$

Find the distance between the vertices $B\left(1,5\right)\mathrm{and}C\left(2,1\right)$.

$$\begin{gathered} BC=\sqrt{{\left(2-1\right)}^2+{\left(1-5\right)}^2} \\ BC=\sqrt{{\left(1\right)}^2+{\left(-4\right)}^2} \\ BC=\sqrt{17} \end{gathered}$$

Find the distance between the vertices $C\left(2,1\right)\mathrm{and}D\left(6,2\right)$.

$$\begin{gathered} CD=\sqrt{{\left(6-2\right)}^2+{\left(2-1\right)}^2} \\ CD=\sqrt{{\left(4\right)}^2+{\left(1\right)}^2} \\ CD=\sqrt{17} \end{gathered}$$

Find the distance between the vertices $A\left(5,6\right)\mathrm{and}D\left(6,2\right)$.

$$\begin{gathered} DA=\sqrt{{\left(6-5\right)}^2+{\left(2-6\right)}^2} \\ DA=\sqrt{{\left(1\right)}^2+{\left(-4\right)}^2} \\ DA=\sqrt{17} \end{gathered}$$

Observe that $AB=BC=CD=DA$.

Hence, all the sides are equal.

Step 2. Determine whether the angle between any two sides is at a right angle.

To find whether the slopes are perpendicular, find that the product of both slopes is $-1$ since the product of slopes is $-1$ if they are perpendicular.

The slope is given by $\mathrm{slope}=\frac{y_2-y_1}{x_2-x_1}$.

The slope of $AB\mathrm{and}BC$ is as follows:

$\begin{array}{rcl}\mathrm{The}\mathrm{slope}\mathrm{of}AB& =& \frac{5-6}{1-5}\\ & \Rightarrow & \mathrm{The}\mathrm{slope}\mathrm{of}AB=\frac{1}{4}\\ \mathrm{The}\mathrm{slope}\mathrm{of}BC& =& \frac{1-5}{2-1}\\ & \Rightarrow & \mathrm{The}\mathrm{slope}\mathrm{of}BC=-4\end{array}$

$$\begin{gathered} \mathrm{Slope}\mathrm{of}AB\times \mathrm{Slope}\mathrm{of}BC=\frac{1}{4}\times \left(-4\right) \\ \mathrm{Slope}\mathrm{of}AB\times \mathrm{Slope}\mathrm{of}BC=-1 \end{gathered}$$

Therefore, it is proved that the angle between the two vertices is at right angles.

Thus, it is proved that all the sides are equal and the angle between the two vertices is at right angles.

Hence, it is proved that $A\left(5,6\right),B\left(1,5\right),C\left(2,1\right),\mathrm{and}D\left(6,2\right)$ are the vertices of a square.