Chapter 6.3, Problem 27PPS

To determine

The graph of each quadrilateral with given vertices and whether the figure forms a parallelogram or not.

Answer to Problem 27PPS

QRST is a parallelogram as one of the pairs of the opposite sides is both, congruent and parallel.

Explanation of Solution

Given:

The given vertices are:

  $Q\left(2,-4\right),R\left(4,3\right),S\left(-3,6\right),T\left(-5,-1\right).$

Calculation:

Parallelogram has one of its pairs of opposite sides as parallel and congruent.

The quadrilateral with given vertices will be:

The slope of line segment QR will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{3-\left(-4\right)}{4-2}\\ m=\frac{7}{2}\end{array}$

The slope of line segment ST will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{-1-6}{-5-\left(-3\right)}\\ m=\frac{-7}{-2}\\ m=\frac{7}{2}\end{array}$

The distance of QR will be:

  $\begin{array}{l}QR=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ QR=\sqrt{{\left(4-2\right)}^2+{\left(3-\left(-4\right)\right)}^2}\\ QR=\sqrt{4+49}\\ QR=\sqrt{53}\end{array}$

The distance of ST will be:

  $\begin{array}{l}ST=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ ST=\sqrt{{\left(-5-\left(-3\right)\right)}^2+{\left(-1-6\right)}^2}\\ ST=\sqrt{4+49}\\ ST=\sqrt{53}\end{array}$

The slope of line QR is equal to the slope of line ST. Therefore, $QR\parallel ST.$

As, QR = ST, so $\overline{QR}\cong \overline{ST.}$

Therefore, by the definition of a parallelogram, QRST is a parallelogram as one of the pairs of the opposite sides is both, congruent and parallel.