Chapter 5.6, Problem 12PSB

a.

To determine

To prove:Quadrilateral PQRSis not a parallelogram.

Explanation of Solution

Given information:InQuadrilateral PQRS, The coordinates of point Pare$\left(-10,7\right)$ .

The coordinates of point Q are $\left(4,3\right)$ .

The coordinates of point Rare$\left(-2,-5\right)$ .

The coordinates of point S are $\left(-16,1\right)$ .

Formula used:

Slope of a line:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=slope\end{array}$

  $(x_1,y_1)$ arecoordinates of first point in the line.

  $(x_2,y_2)$ arecoordinates of second point in the line.

Proof:

  

The coordinates of point Pare$\left(-10,7\right)$ .

The coordinates of point Q are $\left(4,3\right)$ .

The coordinates of point Rare$\left(-2,-5\right)$ .

The coordinates of point S are $\left(-16,1\right)$ .

We now check the slopes of QuadrilateralPQRS.

Slope of $\overline{PS}=\frac{7-1}{-10-\left(-16\right)}=\frac{6}{6}=1$

Slope of $\overline{QR}=\frac{3-\left(-5\right)}{4-\left(-2\right)}=\frac{8}{6}=\frac{4}{3}$

Two lines are parallel if their slopes are equal.

  $\overline{PS}$ is not parallel to $\overline{QR}$ .

Quadrilateral PQRS is not a parallelogram because both pairs of opposite sides have to be parallel.

b.

To determine

To prove: Quadrilateral ABCD is a parallelogram.

Explanation of Solution

Given information:InQuadrilateral PQRS, The coordinates of point Pare$\left(-10,7\right)$ .

The coordinates of point Q are $\left(4,3\right)$ .

The coordinates of point Rare$\left(-2,-5\right)$ .

The coordinates of point S are $\left(-16,1\right)$ .

Formula used:

Slope of a line:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=slope\end{array}$

  $(x_1,y_1)$ arecoordinates of first point in the line.

  $(x_2,y_2)$ arecoordinates of second point in the line.

Proof:

  

The coordinates of point Pare$\left(-10,7\right)$ .

The coordinates of point Q are $\left(4,3\right)$ .

The coordinates of point Rare$\left(-2,-5\right)$ .

The coordinates of point S are $\left(-16,1\right)$ .

We now check the slopes of QuadrilateralABCD.

Slope of $\overline{AD}=\frac{4-\left(-2\right)}{-13-\left(-9\right)}=\frac{6}{-4}=-\frac{3}{2}$

Slope of $\overline{BC}=\frac{5-\left(-1\right)}{-3-1}=\frac{6}{-4}=-\frac{3}{2}$

Two lines are parallel if their slopes are equal.

  $\overline{AD}$ is parallel to $\overline{BC}$ .

Slope of $\overline{AB}=\frac{5-4}{-3-\left(-13\right)}=\frac{1}{10}$

Slope of $\overline{DC}=\frac{-2-\left(-1\right)}{-9-1}=\frac{-1}{-10}=\frac{1}{10}$

Two lines are parallel if their slopes are equal.

  $\overline{AB}$ is parallel to $\overline{DC}$ .

Quadrilateral ABCD is a parallelogram because both pairs of opposite sides are parallel.