Chapter 2.2, Problem 26E

To determine

To show: that the given point coordinates are the vertices of a square.

Answer to Problem 26E

The four sides of a square are equal. The sides of squares are AB, BC, CD, and AD is 5 units.

Explanation of Solution

Given information:

The given point coordinates are $\left(-1,3\right),\left(3,6\right),\left(6,2\right)$ and $\left(2,-1\right)$ .

Calculation:

Let the point coordinates are $\left(-1,3\right),\left(3,6\right),\left(6,2\right)$ and $\left(2,-1\right)$ be A, B, C, and D.

The distance formula is as follows:

  $\text{Distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$  ... (1)

Put the values $\left(-1,3\right)\text{ and}\left(3,6\right)$ and evaluate the distance between them as follows:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(3-\left(-1\right)\right)}^2+{\left(6-3\right)}^2}\\ =\sqrt{{\left(3+1\right)}^2+{\left(3\right)}^2}\\ =\sqrt{{\left(4\right)}^2+{\left(3\right)}^2}\\ =\sqrt{16+9}\\ =\sqrt{25}\\ =5\end{array}$

AB, BC, CD, AD=5

AC, BD=

Put the values $\left(-1,3\right)\text{ and}\left(6,2\right)$ and evaluate the distance between them as follows:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(6-\left(-1\right)\right)}^2+{\left(2-3\right)}^2}\\ =\sqrt{{\left(6+1\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{{\left(7\right)}^2+1}\\ =\sqrt{49+1}\\ =\sqrt{50}\end{array}$

Put the values $\left(-1,3\right)\text{ and}\left(2,-1\right)$ and evaluate the distance between them as follows:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(2-\left(-1\right)\right)}^2+{\left(-1-3\right)}^2}\\ =\sqrt{{\left(2+1\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{{\left(3\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

Put the values $\left(3,6\right)\text{ and}\left(6,2\right)$ and evaluate the distance between them as follows:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(6-3\right)}^2+{\left(2-6\right)}^2}\\ =\sqrt{{\left(3\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

Put the values $\left(3,6\right)\text{ and}\left(2,-1\right)$ and evaluate the distance between them as follows:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(2-3\right)}^2+{\left(-1-6\right)}^2}\\ =\sqrt{{\left(-1\right)}^2+{\left(-7\right)}^2}\\ =\sqrt{1+49}\\ =\sqrt{50}\end{array}$

Put the values $\left(6,2\right)\text{ and}\left(2,-1\right)$ and evaluate the distance between them as follows:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(2-6\right)}^2+{\left(-1-2\right)}^2}\\ =\sqrt{{\left(-4\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{16+9}\\ =\sqrt{25}\\ =5\end{array}$

The two distances AC and BD is $\sqrt{50}$ . The four distances AB, BC, CD, and AD is 5.

The four sides of a square are equal. The sides of squares are AB, BC, CD, and AD; and the diagonals are AC and BD. Thus, the distance between these points coordinates $\left(-1,3\right),\left(3,6\right),\left(6,2\right)$ and $\left(2,-1\right)$ shows the vertex of a square.