Question

Draw a graph for the original figure and its dilated image. Check whether the dilation is a similarity transformation or not.
Given:
The given vertices are
original \(\displaystyle\rightarrow{A}{\left({2},{3}\right)},{B}{\left({0},{1}\right)},{C}{\left({3},{0}\right)}\)
image \(\displaystyle\rightarrow{D}{\left({4},{6}\right)},{F}{\left({0},{2}\right)},{G}{\left({6},{0}\right)}\)

Answer 1

Calculation:
The graph for the points is given below.

Find the distance (using distance formula) of corresponding sides and find the ratio.
\(\displaystyle{A}{B}=\sqrt{{{\left({2}-{0}\right)}^{{2}}+{\left({3}-{1}\right)}^{{2}}}}=\sqrt{{{8}}}={2}\sqrt{{{2}}}\)
\(\displaystyle{D}{F}=\sqrt{{{\left({4}-{0}\right)}^{{2}}+{\left({6}-{2}\right)}^{{2}}}}=\sqrt{{{32}}}={4}\sqrt{{{2}}}\)
\(\displaystyle\frac{{{A}{B}}}{{{D}{F}}}=\frac{{{2}\sqrt{{{2}}}}}{{{4}\sqrt{{{2}}}}}=\frac{{1}}{{2}}\)
\(\displaystyle{A}{C}=\sqrt{{{\left({2}-{3}\right)}^{{2}}+{\left({3}-{0}\right)}^{{2}}}}=\sqrt{{{10}}}\)
\(\displaystyle{D}{G}=\sqrt{{{\left({4}-{6}\right)}^{{2}}+{\left({6}-{0}\right)}^{{2}}}}=\sqrt{{{40}}}={2}\sqrt{{{10}}}\)
\(\displaystyle\frac{{{A}{C}}}{{{D}{G}}}=\frac{{\sqrt{{{10}}}}}{{{2}\sqrt{{{10}}}}}=\frac{{1}}{{2}}\)
\(\displaystyle{B}{C}=\sqrt{{{\left({0}-{3}\right)}^{{2}}+{\left({1}-{0}\right)}^{{2}}}}=\sqrt{{{10}}}\)
\(\displaystyle{F}{G}=\sqrt{{{\left({0}-{6}\right)}^{{2}}+{\left({2}-{0}\right)}^{{2}}}}=\sqrt{{{40}}}={2}\sqrt{{{10}}}\)
\(\displaystyle\frac{{{B}{C}}}{{{F}{G}}}=\frac{{\sqrt{{{10}}}}}{{{2}\sqrt{{{10}}}}}=\frac{{1}}{{2}}\)
From the above result, the lenght of the sides are proportional.
By the use of SSS similarity,
\(\displaystyle\triangle{D}{F}{G}\sim\triangle{A}{B}{C}\).
Hence the dilation is a similarity transformation.