Question

Show that $△ABC$, where A(-2, 0), B(0, 2), and C(2, 0) and $△PQR$ where P(-4, 0), Q(4, 0), R(0, 4) are similar triangles.

Answer 1

To prove that $\Delta ABC\sim \Delta PQR$

we need to show that $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

We know that the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is,

$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

In $\Delta$ABC vertices are A(-2, 0), B(0, 2), and C(2, 0).

$AB=\sqrt{(0--2{)}^2+(2-0{)}^2}=\sqrt{(2{)}^2+(2{)}^2}=\sqrt{8}=2\sqrt{2}$

$BC=\sqrt{(2-0{)}^2+(0-2{)}^2}=\sqrt{(2{)}^2+(2{)}^2}=\sqrt{8}=2\sqrt{2}$

$AC=\sqrt{(2-(-2){)}^2+(0-0{)}^2}=\sqrt{(4{)}^2}=4$

In $\Delta$PQR vertices are P(-4, 0), Q(4, 0) and R(0, 4).

$PQ=\sqrt{(0--4{)}^2+(4-0{)}^2}=\sqrt{(4{)}^2+(4{)}^2}=\sqrt{32}=4\sqrt{2}$

$QR=\sqrt{(4-0{)}^2+(0-4{)}^2}=\sqrt{(4{)}^2+(4{)}^2}=\sqrt{32}=4\sqrt{2}$

$PR=\sqrt{(2-(-2){)}^2+(0-0{)}^2}=\sqrt{(8{)}^2}=8$

$\frac{AB}{PQ}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}\frac{BC}{QR}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}\frac{AC}{PR}=\frac{4}{8}=\frac{1}{2}$

$i.e,\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

$\therefore \Delta ABC\sim \Delta PQR$