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 J(-6,8),K(6,6),L(-2,4),D(-12,16),G(12,12),H(-4,8)

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asked 2021-02-13
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
J(-6,8),K(6,6),L(-2,4),D(-12,16),G(12,12),H(-4,8)

Answers (1)

2021-02-14

Calculation:
The graph for the given points J(-6,8),K(6,6),L(-2,4),D(-12,16),G(12,12),H(-4,8) is given below.
image
Find the distance(using distance formula)of corresponding sides and find the ratio.
\(\displaystyle{D}{H}=\sqrt{{{\left[-{12}-{\left(-{4}\right)}\right]}^{{2}}+{\left[{16}-{8}\right]}^{{2}}}}=\sqrt{{{128}}}={8}\sqrt{{{2}}}\)
\(\displaystyle{J}{L}=\sqrt{{{\left[-{6}-{\left(-{2}\right)}\right]}^{{2}}+{\left({8}-{4}\right)}^{{2}}}}=\sqrt{{{32}}}={4}\sqrt{{{2}}}\)
\(\displaystyle\frac{{{D}{H}}}{{{J}{L}}}=\frac{{{8}\sqrt{{{2}}}}}{{{4}\sqrt{{{2}}}}}={2}\)
\(\displaystyle{G}{H}=\sqrt{{{\left[-{12}-{\left(-{4}\right)}\right]}^{{2}}+{\left({12}-{8}\right)}^{{2}}}}=\sqrt{{{272}}}={4}\sqrt{{{17}}}\)
\(\displaystyle{K}{L}=\sqrt{{{\left[{6}-{\left(-{2}\right)}\right]}^{{2}}+{\left({6}-{4}\right)}^{{2}}}}=\sqrt{{{68}}}={2}\sqrt{{{17}}}\)
\(\displaystyle\frac{{{G}{H}}}{{{K}{L}}}=\frac{{{4}\sqrt{{{17}}}}}{{{2}\sqrt{{{17}}}}}={2}\)
\(\displaystyle{D}{G}=\sqrt{{{\left[-{12}-{12}\right)}}}^{{2}}+{\left({16}-{12}\right)}^{{2}}{)}=\sqrt{{{592}}}={4}\sqrt{{{37}}}\)
\(\displaystyle{J}{K}=\sqrt{{{\left[-{6}-{6}\right]}^{{2}}+{\left({8}-{6}\right)}^{{2}}}}=\sqrt{{{148}}}={2}\sqrt{{{37}}}\)
\(\displaystyle\frac{{{D}{G}}}{{{J}{K}}}=\frac{{{4}\sqrt{{{37}}}}}{{{2}\sqrt{{{37}}}}}={2}\)
From the above result, the lenght of the sides are proportional and the angles between them are congruent.
By the use of SSS similarity,
\(\displaystyle\triangle{J}{L}{K}\sim\triangle{D}{H}{G}\).
Hence the dilation is a similarity transformation.

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