Chapter 13.5, Problem 18WE

a.

To determine

To find: Graph the points $A(-5,0),B(3,2),C(5,6)\text{ and }D(-3,4)$ . Then show that $ABCD$ is a parallelogram by two different methods.

Explanation of Solution

Given information:Points $A(-5,0),B(3,2),C(5,6)\text{ and }D(-3,4)$ .

Graph:

  

b.

To determine

To find:Show that one pair of opposite sides are both congruent and parallel.

Explanation of Solution

Given information:Points $A(-5,0),B(3,2),C(5,6)\text{ and }D(-3,4)$ .

Formula used: The distance between two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is given by:

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

Theslope of a line passing through point $A(x_1,y_1)$ and $B(x_2,y_2)$ is $\frac{y_2-y_1}{x_2-x_1}$ .

Concept used: In a quadrilateral if one pair of opposite sides are both congruent and parallel then it is a parallelogram.

Calculation:

Here, sides AB and CD are opposite sides.

  $\begin{array}{l}{d}_{AB}=\sqrt{{\left(3-(-5)\right)}^2+{\left(2-0\right)}^2}\\ \Rightarrow d_{AB}=\sqrt{{\left(8\right)}^2+{\left(2\right)}^2}\\ \Rightarrow d_{AB}=\sqrt{64+4}\\ \Rightarrow d_{AB}=\sqrt{68}\\ \Rightarrow d_{AB}=2\sqrt{17}\text{ units}\\ d_{CD}=\sqrt{{\left(-3-5\right)}^2+{\left(4-6\right)}^2}\\ \Rightarrow d_{CD}=\sqrt{{\left(-8\right)}^2+{\left(-2\right)}^2}\\ \Rightarrow d_{CD}=\sqrt{64+4}\\ \Rightarrow d_{CD}=\sqrt{68}\\ \Rightarrow d_{CD}=2\sqrt{17}\text{ units}\end{array}$

Since, length of AB and CD are equal.

So, sides AB and CD are congruent.

  $\begin{array}{l}\text{Slope of }\overline{AB}=\frac{2-0}{3-(-5)}\\ \Rightarrow \text{Slope of }\overline{AB}=\frac{2}{8}\\ \Rightarrow \text{Slope of }\overline{AB}=\frac{1}{4}\\ \text{Slope of }\overline{CD}=\frac{4-6}{-3-5}\\ \Rightarrow \text{Slope of }\overline{CD}=\frac{-2}{-8}\\ \Rightarrow \text{Slope of }\overline{CD}=\frac{1}{4}\end{array}$

Since, slopes of $\overline{AB}$ and slope of $\overline{CD}$ are equal.

So, sides AB and CD are parallel.

Since, one pair of opposite sides are both congruent and parallel.

Hence, $ABCD$ is a parallelogram.

c.

To determine

To find: Show that the diagonals bisects each other (have the same midpoints).

Explanation of Solution

Given information:points $A(-5,0),B(3,2),C(5,6)\text{ and }D(-3,4)$ .

Formula Used:If points are $A(x_1,y_1)\text{ and }B(x_2,y_2)$ then coordinates of midpoint of $\overline{AB}$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .

Calculation:

Here, AC and BD are diagonals of $ABCD$ .

Let midpoints of $\overline{AC}$ and $\overline{BD}$ are S and T.

Coordinates of S is $\left(\frac{-5+5}{2},\frac{0+6}{2}\right)\equiv \left(0,3\right)$

Coordinates of T is $\left(\frac{3-3}{2},\frac{2+4}{2}\right)\equiv \left(0,3\right)$

Since both of the midpoints is $\left(0,3\right)$

So, $\overline{AC}$ and $\overline{BD}$ have the same midpoint.

Since, both diagonals has same midpoint.

It means that both diagonals intersect each other at $\left(0,3\right)$ which is midpoint.

So, diagonals of $ABCD$ bisects each other.

If diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Hence, $ABCD$ is a parallelogram.