Chapter 1, Problem 15T

1. (a) Plot the points P(0, 3), Q(3, 0), and R(6, 3) in the coordinate plane. Where must the point S be located so that PQRS is a square?
2. (b) Find the area of PQRS.

To determine

To evaluate: The point S of a square PQRS and plot points in the coordinate plane.

Answer to Problem 15T

The point S is $\left(3,6\right)$.

Explanation of Solution

Given:

The coordinates of points of square PQRS is $P\left(0,3\right)$, $Q\left(3,0\right)$ and $R\left(6,3\right)$.

Formula used:

The distance formula of two points $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ is,

$d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ (1)

Where, $d$ is the distance between two points and $x_1$, $x_2$, $y_1$ and $y_2$ are coordinates of two points.

Calculation:

Let the coordinates of points be $S\left(x,y\right)$.

Find the distance between the points $P\left(0,3\right)$ and $Q\left(3,0\right)$.

Substitute 3 for $x_2$, 0 for $x_1$, 0 for $y_2$ and 3 for $y_1$ in formula (1).

$\begin{array}{c}d\left(P,Q\right)=\sqrt{{\left(3-0\right)}^2+{\left(0-3\right)}^2}\\ =\sqrt{9+9}\\ =\sqrt{18}\\ =3\sqrt{2}\end{array}$

The distance between two points $R\left(6,3\right)$ and $S\left(x,y\right)$ give another side of square.

Substitute $x$ for $x_2$, 6 for $x_1$, $y$ for $y_2$ and 3 for $y_1$ in formula (1).

$\begin{array}{c}d\left(R,S\right)=\sqrt{{\left(x-6\right)}^2+{\left(y-3\right)}^2}\\ =\sqrt{x^2-12x+36+y^2-6y+9}\\ =\sqrt{x^2+y^2-12x-6y+45}\end{array}$

Since all sides of square are equal in length, $d\left(P,Q\right)=d\left(R,S\right)$.

Substitute $3\sqrt{2}$ for $d\left(R,S\right)$ in $d\left(P,Q\right)=d\left(R,S\right)$.

$\begin{array}{c}{\left(\sqrt{x^2+y^2-12x-6y+45}\right)}^2=\left(3{\sqrt{2}}^2\right)\\ x^2+y^2-12x-6y+45=18\\ x^2+y^2-12x-6y+27=0\end{array}$

From the above expression the equation is,

$x^2+y^2-12x-6y+27=0$ (1)

Substitute $\sqrt{18}$ for $d\left(P,S\right)$ to get the distance between point $P\left(0,3\right)$ and $S\left(x,y\right)$.

$\begin{array}{c}\sqrt{{\left(x-0\right)}^2+{\left(y-3\right)}^2}=d\left(P,S\right)\\ x^2+y^2-6y+9={\left(\sqrt{18}\right)}^2\\ x^2+y^2-6y-9=0\end{array}$

The equation from above expression is,

$x^2+y^2-6y-9=0$ (2)

Subtract equation (2) from equation (1) to get the values of x,

$\begin{array}{r}{x}^2+y^2-12x-6y+27=0\\ \underset{\_}{x^2+y^2-6y-9=0}\\ -12x+36=0\\ x=3\end{array}$

Substitute 3 for x in equation (2) to get value of y.

$\begin{array}{l}{\left(3\right)}^2+y^2-6y-9=0\\ y^2-6y=0\\ y\left(y-6\right)=0\\ y=0,6\end{array}$

These values of x- and y-coordinates gives the point S.

Since the point $\left(3,0\right)$ exists already in square, so take $y=6$ in point $S\left(x,y\right)$.

Thus, the coordinates of point S is $\left(3,6\right)$

Now the point $S\left(3,6\right)$ is obtained.

In point $P\left(0,3\right)$, x-coordinate is zero, so it is lies in y-axis similarly point $Q\left(3,0\right)$ lies on x-axis and points $R\left(6,3\right)$, $S\left(3,6\right)$ both coordinates are positive so it is lies in I Quadrant.

The coordinate plane of points of square is shown below in Figure (1).

Figure (1)

From Figure (1), it is observed that the coordinates of the square PQRS are $P\left(0,3\right)$, $Q\left(3,0\right)$ and $R\left(6,3\right)$ and $S\left(3,6\right)$.

To determine

To find: The area of square PQRS.

Answer to Problem 15T

The area of square PQRS is 18.

Explanation of Solution

Given:

The coordinates of points of square PQRS is $P\left(0,3\right)$, $Q\left(3,0\right)$, $R\left(6,3\right)$.

Calculation:

The formula of area of square is,

$\text{area}={\left(\text{side}\right)}^2$

From part (a) the side of square PQRS is $3\sqrt{2}$.

Substitute $3\sqrt{2}$ for side in the formula of area of square.

$\begin{array}{c}\text{Area}={\left(3\sqrt{2}\right)}^2\\ =9\left(2\right)\\ =18\end{array}$

Thus, the area of square PQRS is 18.