Chapter 6.5, Problem 19PPS

To determine

Whether JKLM is a rhombus, a rectangle or a square.

Answer to Problem 19PPS

QRST is a rhombus.

Explanation of Solution

Given:

The vertices are:

  $J\left(-4,-1\right),K\left(1,-1\right),L\left(4,3\right),M\left(-1,3\right).$

Calculation:

Plot the vertices on the graph:

According to Rectangle Theorem, if the diagonals of a parallelogram are congruent, then it is a rectangle.

According to Rhombi Theorem, if the diagonals of a parallelogram are perpendicular, then it is a rhombus.

According to Square Theorem, if the quadrilateral is both, a rectangle and a rhombus, then it is a square.

Consider the diagonals of JKLM:

Finding the length of JL:

  $\begin{array}{l}JL=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ JL=\sqrt{{\left(4-\left(-4\right)\right)}^2+{\left(3-\left(-1\right)\right)}^2}\\ JL=\sqrt{64+16}\\ JL=\sqrt{80}\end{array}$

Finding the length of KM:

  $\begin{array}{l}KM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ KM=\sqrt{{\left(-1-1\right)}^2+{\left(3-\left(-1\right)\right)}^2}\\ KM=\sqrt{4+16}\\ KM=\sqrt{20}\end{array}$

Both of these diagonals are not congruent, hence, QRST is not a rectangle.

Finding the slope of JL:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{3-\left(-1\right)}{4-\left(-4\right)}\\ m=\frac{1}{2}\end{array}$

Finding the slope of KM:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{3-\left(-1\right)}{-1-1}\\ m=\frac{4}{-2}\\ m=-2\end{array}$

As, $\frac{1}{2}\left(-2\right)=-1$ , the diagonals are perpendicular.

Hence, QRST is a rhombus.