Chapter 4.7, Problem 3CYP

To determine

To identify: The type of transformation and show that it is a congruence transformation.

Answer to Problem 3CYP

The given congruence shows a rotation.

Explanation of Solution

Given information: The vertices of $\Delta JKL$ are $J\left(-2,2\right),K\left(-8,5\right)\text{and}L\left(-4,6\right)$ .

The vertices of $\Delta PQR$ are $P\left(2,-2\right),Q\left(8,-5\right)\text{and}R\left(4,-6\right)$ .

Formula used: $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation: The graph of the original figure and its image,

  

The transformation appears to be a rotation.

Now, finding the measures of the side of each triangle,

  $\begin{array}{l}JK=\sqrt{{\left(-8+2\right)}^2+{\left(5-2\right)}^2}\\ =\sqrt{36+9}\\ =\sqrt{45}\end{array}$$\begin{array}{l}PQ=\sqrt{{\left(8-2\right)}^2+{\left(-5+2\right)}^2}\\ =\sqrt{36+9}\\ =\sqrt{45}\end{array}$

  $\begin{array}{l}KL=\sqrt{{\left(-4+8\right)}^2+{\left(6-5\right)}^2}\\ =\sqrt{16+1}\\ =\sqrt{17}\end{array}$$\begin{array}{l}QR=\sqrt{{\left(4-8\right)}^2+{\left(-6+5\right)}^2}\\ =\sqrt{16+1}\\ =\sqrt{17}\end{array}$

  $\begin{array}{l}JL=\sqrt{{\left(-4+2\right)}^2+{\left(6-2\right)}^2}\\ =\sqrt{4+16}\\ =\sqrt{20}\end{array}$$\begin{array}{l}PR=\sqrt{{\left(4-2\right)}^2+{\left(-6+2\right)}^2}\\ =\sqrt{4+16}\\ =\sqrt{20}\end{array}$

Here, $JK=PQ,KL=QR\text{and}JL=PR$ and $\overline{JK}=\overline{PQ},\overline{KL}=\overline{QR}\text{and}\overline{JL}=\overline{PR}$

By side-side-side congruence postulate, $\Delta JKL\cong \Delta PQR$