Chapter 7.6, Problem 3ACYP

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 3ACYP

The dilation is a similarity transformation.

Explanation of Solution

Given:

The given vertices are

original $\to A\left(2,3\right),B\left(0,1\right),C\left(3,0\right)$

image $\to D\left(4,6\right),F\left(0,2\right),G\left(6,0\right)$

Calculation:

The graph for the points is given below.

  

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

  $\begin{array}{l}AB=\sqrt{{\left(2-0\right)}^2+{\left(3-1\right)}^2}=\sqrt{8}=2\sqrt{2}\\ DF=\sqrt{{\left(4-0\right)}^2+{\left(6-2\right)}^2}=\sqrt{32}=4\sqrt{2}\end{array}$

  $\frac{AB}{DF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}$

  $\begin{array}{l}AC=\sqrt{{\left(2-3\right)}^2+{\left(3-0\right)}^2}=\sqrt{10}\\ DG=\sqrt{{\left(4-6\right)}^2+{\left(6-0\right)}^2}=\sqrt{40}=2\sqrt{10}\end{array}$

  $\frac{AC}{DG}=\frac{\sqrt{10}}{2\sqrt{10}}=\frac{1}{2}$

  $\begin{array}{l}BC=\sqrt{{\left(0-3\right)}^2+{\left(1-0\right)}^2}=\sqrt{10}\\ FG=\sqrt{{\left(0-6\right)}^2+{\left(2-0\right)}^2}=\sqrt{40}=2\sqrt{10}\end{array}$

  $\frac{BC}{FG}=\frac{\sqrt{10}}{2\sqrt{10}}=\frac{1}{2}$

From the above result, the length of the sides are proportional.

By the use of $SSS$ similarity,

  $\Delta DFG\sim \Delta ABC$ .

Hence the dilation is a similarity transformation.