Chapter 13.5, Problem 21WE

a.

To determine

To find:The coordinates of the midpoints $J,K,L,\text{ and }M$ .

Answer to Problem 21WE

Coordinates of points are $J\left(-0.5,1.5\right),K(3,6),L(8.5,4.5),\text{ and }M(5,0)$ .

Explanation of Solution

Given information:In figure,points $O(0,0),P(-1,3),Q(7,9),\text{ and }R(10,0)$ and $J,K,L,\text{ and }M$ are midpoints of $\overline{OP},\overline{PQ},\overline{QR}\text{ and }\overline{OR}$ respectively.

Formula Used:The midpoint of segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is the point $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .

Calculation:

Since, $J$ is the midpoint of $\overline{OP}$ .

So, coordinates of point $J$ is $\left(\frac{0+(-1)}{2},\frac{0+3}{2}\right)\equiv \left(-0.5,1.5\right)$

Since, $K$ is the midpoint of $\overline{PQ}$ .

So, coordinates of point $K$ is $\left(\frac{-1+7}{2},\frac{3+9}{2}\right)\equiv \left(3,6\right)$

Since, $L$ is the midpoint of $\overline{QR}$ .

So, coordinates of point $L$ is $\left(\frac{7+10}{2},\frac{9+0}{2}\right)\equiv \left(8.5,4.5\right)$

Since, $M$ is the midpoint of $\overline{OR}$ .

So, coordinates of point $M$ is $\left(\frac{0+10}{2},\frac{0+0}{2}\right)\equiv \left(5,0\right)$

Hence, coordinates of points are $J\left(-0.5,1.5\right),K(3,6),L(8.5,4.5),\text{ and }M(5,0)$ .

b.

To determine

To find:What kind of figure is $JKLM$ ? Prove it.

Answer to Problem 21WE

  $JKLM$ is a parallelogram.

Explanation of Solution

Given information:Obtained from part a: $J\left(-0.5,1.5\right),K(3,6),L(8.5,4.5),\text{ and }M(5,0)$

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

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

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 $JK$ and $LM$ are opposite sides.

  $\begin{array}{l}JK=\sqrt{{\left(3-(-0.5)\right)}^2+{\left(6-1.5\right)}^2}\\ \Rightarrow JK=\sqrt{{\left(3.5\right)}^2+{\left(4.5\right)}^2}\\ \Rightarrow JK=\sqrt{12.25+20.25}\\ \Rightarrow JK=\sqrt{32.5}\\ LM=\sqrt{{\left(5-8.5\right)}^2+{\left(0-4.5\right)}^2}\\ \Rightarrow LM=\sqrt{{\left(-3.5\right)}^2+{\left(-4.5\right)}^2}\\ \Rightarrow LM=\sqrt{12.25+20.25}\\ \Rightarrow LM=\sqrt{32.5}\end{array}$

Since, length of $JK$ and $LM$ are equal.

So, sides $JK$ and $LM$ are congruent.

  $\begin{array}{l}\text{Slope of }\overline{JK}=\frac{6-1.5}{3-(-0.5)}\\ \Rightarrow \text{Slope of }\overline{JK}=\frac{4.5}{3.5}\\ \text{Slope of }\overline{LM}=\frac{0-4.5}{5-8.5}\\ \Rightarrow \text{Slope of }\overline{LM}=\frac{-4.5}{-3.5}\\ \Rightarrow \text{Slope of }\overline{LM}=\frac{4.5}{3.5}\end{array}$

Since, slopes of $\overline{JK}$ and slope of $\overline{LM}$ are equal.

So, sides $JK$ and $LM$ are parallel.

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

Hence, $JKLM$ is a parallelogram.