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Show that $\triangle ABC$, where $A (-2, 0), B (2, 0), C (0, 2)$ and $\triangle PQR$, where $P (-4, 0), Q (4, 0), R (0, 4)$ are similar.
Given:
$A (-2, 0), B (2, 0), C (0, 2)$ are vertices of $\triangle ABC$ and $P (-4, 0), Q (4, 0), R (0, 4)$ are vertices of $\triangle PQR$.
To do:
We have to show that $\triangle ABC$ and $\triangle PQR$ are similar.
Solution:
We know that,
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
In $\triangle ABC$,
\( A B=\sqrt{(2+2)^{2}+(0-0)^{2}} \)
\( =\sqrt{16} \)
\( =4 \) units
\( B C=\sqrt{(0-2)^{2}+(2-0)^{2}} \)
\( =\sqrt{4+4} \)
\( =\sqrt{8} \)
\( =2 \sqrt{2} \) units
\( C A=\sqrt{(-2-0)^{2}+(0-2)^{2}} \)
\( =\sqrt{4+4} \)
\( =\sqrt{8} \)
\( =2 \sqrt{2} \) units
In $\triangle PQR$,
\( P Q=\sqrt{(4+4)^{2}+(0-0)^{2}} \)
\( =\sqrt{64} \)
\( =8 \) units
\( QR=\sqrt{(0-4)^{2}+(4-0)^{2}} \)
\( =\sqrt{32} \)
\( =4 \sqrt{2} \) units
\( RP=\sqrt{(0+4)^{2}+(4-0)^{2}} \)
\( =\sqrt{16+16} \)
\( =\sqrt{32} \)
\( =4 \sqrt{2} \) units
Here,
\( \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{4}{8}=\frac{1}{2} \)
\( \frac{\mathrm{BC}}{\mathrm{QR}}=\frac{2 \sqrt{2}}{4 \sqrt{2}}=\frac{1}{2} \)
\( \frac{\mathrm{CA}}{\mathrm{PQ}}=\frac{2 \sqrt{2}}{4 \sqrt{2}}=\frac{1}{2} \)
Therefore,
\( \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{RP}} \)
This implies, by SSS similarity,
\( \triangle \mathrm{ABC} \sim\ \triangle \mathrm{PQR} \)
Hence proved.
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