Question

Show that the points $A\left(1,2\right)$, $B\left(5,4\right)$, $C\left(3,8\right)$ and $D\left(-1,6\right)$ are the vertices of a square.

Answer 1

Step 1: Determine the lengths of the sides of the square:

The given coordinates of the points are $A\left(1,2\right)$, $B\left(5,4\right)$, $C\left(3,8\right)$ and $D\left(-1,6\right)$.

The distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ can be given by $\sqrt{{\left(x_1-x_2\right)}^{\mathbf{2}}\mathbf{+}{\left(y_1-y_2\right)}^{\mathbf{2}}}$.

The length of side, $AB=\sqrt{{\left(1-5\right)}^2+{\left(2-4\right)}^2}$.

$$\begin{gathered} =\sqrt{{\left(-4\right)}^2+{\left(-2\right)}^2} \\ =\sqrt{16+4} \\ =\sqrt{20} \end{gathered}$$

$\Rightarrow AB=2\sqrt{5}$

The length of side, $BC=\sqrt{{\left(5-3\right)}^2+{\left(4-8\right)}^2}$.

$$\begin{gathered} =\sqrt{{\left(2\right)}^2+{\left(-4\right)}^2} \\ =\sqrt{4+16} \\ =\sqrt{20} \end{gathered}$$

$\Rightarrow BC=2\sqrt{5}$

The length of side, $CD=\sqrt{{\left(3+1\right)}^2+{\left(8-6\right)}^2}$.

$$\begin{gathered} =\sqrt{{\left(4\right)}^2+{\left(2\right)}^2} \\ =\sqrt{16+4} \\ =\sqrt{20} \end{gathered}$$

$\Rightarrow CD=2\sqrt{5}$

The length of side, $DA=\sqrt{{\left(-1-1\right)}^2+{\left(6-2\right)}^2}$.

$$\begin{gathered} =\sqrt{{\left(-2\right)}^2+{\left(4\right)}^2} \\ =\sqrt{4+16} \\ =\sqrt{20} \end{gathered}$$

$\Rightarrow DA=2\sqrt{5}$

Thus, $AB=BC=CD=DA=2\sqrt{5}$

The lengths of sides are equal in square as well as rhombus, so it can be either a square or rhombus ….$\left[1\right]$

Step 2: Determine the lengths of diagonals:

The length of diagonal, $AC=\sqrt{{\left(1-3\right)}^2+{\left(2-8\right)}^2}$.

$$\begin{gathered} =\sqrt{{\left(-2\right)}^2+{\left(-6\right)}^2} \\ =\sqrt{4+36} \\ =\sqrt{40} \end{gathered}$$

$\Rightarrow AC=\sqrt{40}$

The length of diagonal, $BD=\sqrt{{\left(5+1\right)}^2+{\left(4-6\right)}^2}$.

$$\begin{gathered} =\sqrt{{\left(6\right)}^2+{\left(-2\right)}^2} \\ =\sqrt{36+4} \\ =\sqrt{40} \end{gathered}$$

$\Rightarrow BD=\sqrt{40}$

Thus, $AC=BD=\sqrt{40}$

The lengths of diagonals are equal in square as well as rectangle, so it can be either a square or rectangle ….$\left[2\right]$

From $\left[1\right]$ and $\left[2\right]$ we can conclude that the given points form a square

Hence, the points $A\left(1,2\right)$, $B\left(5,4\right)$, $C\left(3,8\right)$ and $D\left(-1,6\right)$ are the vertices of a square.