Chapter 5.6, Problem 5PSA

To determine

To prove: Quadrilateral PQRS is a parallelogram.

Explanation of Solution

Given information:Coordinates of the end points of quadrilateral are $P\left(-10,6\right)$ , $Q\left(2,6\right)$ , $R\left(5,-4\right)$ and $S\left(-7,-4\right)$ .

Formula used:If $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are the coordinates of end points of any line segment then,

(a) Slope of line $\left(m\right)=\left(\frac{y_2-y_1}{x_2-x_1}\right)$ …………………………… (i)

(b) Length of the line $\left(l\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ ………… (ii)

Proof: Slope of the line joining points $P\left(-10,6\right)$ and $S\left(-7,-4\right)$ by using equation (i)

  $\begin{array}{l}{m}_1=\left(\frac{-4-6}{-7-\left(-10\right)}\right)\\ m_1=\left(\frac{-10}{-7+10}\right)\\ m_1=\left(\frac{-10}{3}\right)\end{array}$

Now, slope of the line joining points $Q\left(2,6\right)$ and $R\left(5,-4\right)$ by using equation (i)

  $\begin{array}{l}{m}_2=\left(\frac{-4-6}{5-2}\right)\\ m_2=\left(\frac{-10}{3}\right)\end{array}$

So, we have $m_1=m_2$

i.e., slopes of lines PS and QR is equal which means $PS\parallel QR$ .

Let us calculate the lengths of line PS and QR by using equation (ii)

Length of the line PS $\left(l_1\right)=\sqrt{{\left(-7-\left(-10\right)\right)}^2+{\left(-4-6\right)}^2}$

  $\begin{array}{l}{l}_1=\sqrt{{\left(-7+10\right)}^2+{\left(-4-6\right)}^2}\\ l_1=\sqrt{{\left(3\right)}^2+{\left(-10\right)}^2}\\ l_1=\sqrt{9+100}\\ l_1=\sqrt{109}\text{ units}\end{array}$

Similarly, length of the line QR$\left(l_2\right)=\sqrt{{\left(5-2\right)}^2+{\left(-4-6\right)}^2}$

  $\begin{array}{l}{l}_2=\sqrt{{\left(3\right)}^2+{\left(-10\right)}^2}\\ l_2=\sqrt{9+100}\\ l_2=\sqrt{109}\text{ units}\end{array}$

So, we have $l_1=l_2$

i.e., length of lines PS and QRarecongruent to each other.

Thus, $PS\cong QR$ and $PS\parallel QR$

Hence, a pair of opposite sides are congruent and parallel to each other so the given quadrilateral PQRSis a parallelogram.