Chapter 1.8, Problem 22E

To determine

To calculate: The location of the point S so that quadrilateral PQRS is a square and area of the square along with figure.

Answer to Problem 22E

The coordinates of point S so that PQRS forms a square is $S\left(0,-4\right)$ and its area is $50\text{ sq}\text{. units}$ .

Explanation of Solution

Given information:

The points $P\left(5,1\right),Q\left(0,6\right)\text{ and }R\left(-5,1\right)$ .

Formula used:

A square is a special kind of quadrilateral having equal lengths of its sides and diagonals.

Distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

Area of a square with its side length x is the square of its side, which is mathematically written as,

  $\begin{array}{c}\text{Area}={\left(side\right)}^2\\ =x^2\end{array}$

Calculation:

Consider the provided vertices $P\left(5,1\right),Q\left(0,6\right)\text{ and }R\left(-5,1\right)$ .

By plotting the given points on the coordinate plane, we get the following figure,

  

Recall that a square is a special kind of quadrilateral having equal lengths of its sides and diagonals.

So, to find the coordinates of S such that PQRS forms a square, the lengths of its diagonals i.e. PR and QS must be equal.

  $\overline{\text{PR}}\text{ = }\overline{\text{QS}}$

For the coordinates to form a square, S must lie on y-axis, so, assume the coordinates of S as $\left(0,y\right)$ .

Now, apply distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ which iss calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

  $\begin{array}{c}\overline{\text{PR}}\text{ = }\overline{\text{QS}}\\ \sqrt{{\left(5-\left(-5\right)\right)}^2+{\left(1-1\right)}^2}=\sqrt{{\left(0-0\right)}^2+{\left(6-y\right)}^2}\\ \sqrt{{10}^2}=\sqrt{{\left(6-y\right)}^2}\\ 10=6-y\end{array}$

Simplify it further as,

  $\begin{array}{c}10=6-y\\ y=6-10\\ y=-4\end{array}$

Therefore, the fourth vertex of the square PQRS is at point $S\left(0,-4\right)$ .

Now, the length of side of square is calculated as,

  $\begin{array}{c}\text{PQ}=\sqrt{{\left(0-5\right)}^2+{\left(6-1\right)}^2}\\ =\sqrt{5^2+5^2}\\ =\sqrt{2\cdot 5^2}\\ =5\sqrt{2}\end{array}$

Recall that area of a square with its side length x is the square of its side, which is mathematically written as,

  $\begin{array}{c}\text{Area}={\left(side\right)}^2\\ =x^2\end{array}$

So, area of square PQRS is calculated as,

  $\begin{array}{c}\text{Area}={\left(side\right)}^2\\ ={\left(5\sqrt{2}\right)}^2\\ =25\cdot 2\\ =50\text{ sq}\text{. units}\end{array}$

Thus, the coordinates of point S so that PQRS forms a square is $S\left(0,-4\right)$ and its area is $50\text{ sq}\text{. units}$ .