Chapter 7.6, Problem 16PPS

To determine

Draw a graph for the original figure and its dilated image. Check whether the dilation is a similarity transformation or not.

Answer to Problem 16PPS

The dilation is a similarity transformation.

Explanation of Solution

Given:

The given vertices are

  $V\left(-3,4\right),W\left(-5,0\right),X\left(1,2\right),Y\left(-6,-2\right),Z\left(3,1\right)$

Calculation:

The graph for the given points $V\left(-3,4\right),W\left(-5,0\right),X\left(1,2\right),Y\left(-6,-2\right),Z\left(3,1\right)$ is given below.

  

Find the distance (using distance formula) of corresponding sides and find the ratio.

  $\begin{array}{l}VY=\sqrt{{\left[-3-\left(-6\right)\right]}^2+{\left[4-\left(-2\right)\right]}^2}=\sqrt{45}=3\sqrt{5}\\ VW=\sqrt{{\left[-3-\left(-5\right)\right]}^2+{\left(4-0\right)}^2}=\sqrt{20}=2\sqrt{5}\end{array}$

  $\frac{VW}{VY}=\frac{2\sqrt{5}}{3\sqrt{5}}=\frac{2}{3}$

  $\begin{array}{l}VZ=\sqrt{{\left(-3-3\right)}^2+{\left(4-1\right)}^2}=\sqrt{45}=3\sqrt{5}\\ VX=\sqrt{{\left(-3-1\right)}^2+{\left(4-2\right)}^2}=\sqrt{20}=2\sqrt{5}\end{array}$

  $\frac{VX}{VZ}=\frac{2\sqrt{5}}{3\sqrt{5}}=\frac{2}{3}$

  $\angle V\cong \angle V$

From the above result, the length of two sides are proportional and the angles between them are congruent.

By the use of $SAS$ similarity,

  $\Delta ADE\sim \Delta ABC$ .

Hence the dilation is a similarity transformation.