Question

Show that ΔABC, where A(–2, 0), B(2, 0), C(0, 2) and ΔPQR, where P(–4, 0), Q(4, 0), R(0, 4) are similar triangles.

Answer 1

In ΔABC, the coordinates of the vertices are A(–2, 0), B(2, 0), C(0, 2). 

\[AB = \sqrt{\left( 2 + 2 \right)^2 + \left( 0 - 0 \right)^2} = 4\]
\[BC = \sqrt{\left( 0 - 2 \right)^2 + \left( 2 - 0 \right)^2} = \sqrt{8} = 2\sqrt{2}\]
\[CA = \sqrt{\left( 0 + 2 \right)^2 + \left( 2 - 0 \right)^2} = \sqrt{8} = 2\sqrt{2}\]

In ΔPQR, the coordinates of the vertices are P(–4, 0), Q(4, 0), R(0, 4).

\[PQ = \sqrt{\left( 4 + 4 \right)^2 + \left( 0 - 0 \right)^2} = 8\]
\[QR = \sqrt{\left( 0 - 4 \right)^2 + \left( 4 - 0 \right)^2} = 4\sqrt{2}\]
\[PR = \sqrt{\left( 0 + 4 \right)^2 + \left( 4 - 0 \right)^2} = 4\sqrt{2}\]

Now, for ΔABC and ΔPQR to be similar, the corresponding sides should be proportional.

\[So, \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{PR}\]
\[ \Rightarrow \frac{4}{8} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}\]

Thus, ΔABC is similar to ΔPQR.