Chapter 5.7, Problem 14PSB

To determine

To find: Thename of a quadrilateral whose four vertices are given.

Answer to Problem 14PSB

The quadrilateral is a rectangle.

Explanation of Solution

Given Information:

Vertices of the quadrilateral are $\left(-7,2\right),\left(2,8\right),\left(6,2\right)\text{and}\left(-3,-4\right)$

Formula used:

Distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is given by

  $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

Slope of a line joining two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$

Calculation:

Let the four vertices of the quadrilateral be $A\left(-7,2\right),B\left(2,8\right),C\left(6,2\right)\text{and}D\left(-3,-4\right)$

Using Distance formula, we have

Distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

  $\therefore AB=\sqrt{{\left(-7-2\right)}^2+{\left(2-8\right)}^2}=\sqrt{81+36}=\sqrt{117}$

  $\begin{array}{l}BC=\sqrt{{\left(2-6\right)}^2+{\left(8-2\right)}^2}=\sqrt{16+36}=\sqrt{52}\\ CD=\sqrt{{\left(6+3\right)}^2+{\left(2+4\right)}^2}=\sqrt{81+36}=\sqrt{117}=AB\\ DA=\sqrt{{\left(-3+7\right)}^2+{\left(-4-2\right)}^2}=\sqrt{16+36}=\sqrt{52}=BC\end{array}$

So we get that the opposite sides are equal.

Now, let’s find the slope of the sides $AB,BC,CD\text{and}DA$

Slope of $AB=\frac{8-2}{2+7}=\frac{2}{3}$

Slope of $BC=\frac{2-8}{6-2}=\frac{-3}{2}$

Slope of $CD=\frac{-4-2}{-3-6}=\frac{2}{3}$

Slope of $DA=\frac{2+4}{-7+3}=-\frac{3}{2}$

Therefore, we get

Slope of $AB=$ Slope of $CD$ … (i)

Slope of $BC=$ Slope of $DA$ … (ii)

We know that if the slopes of two lines are equal then they are parallel, then we have

  $AB\parallel CD\text{and}BC\parallel DA$

Also, Slope of $AB=-\frac{1}{\text{Slope}\text{of}BC}$ … (iii)

And, Slope of $CD=-\frac{1}{\text{Slope}\text{of}DA}$ … (iv)

We know that if the slope of a line is equal to the negative reciprocal of the other line then they are perpendicular

  $\Rightarrow AB\perp BC,AB\perp DA,CD\perp BC\text{and}CD\perp DA$

Thus, the opposite sides are equal and parallel. Also, all the angles are right angles.

Hence, it is a rectangle.