Question

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

Answer 1

In $△ABC,$

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

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

$=\sqrt{(2+2{)}^2}$

$=\sqrt{(4{)}^2}$

$=4units$

$\therefore$ $AB=4units$

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

$=\sqrt{(2{)}^2+(2{)}^2}$

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}units$

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

$=\sqrt{4+4}$

$=2\sqrt{2}units$

In $△$PQR$,

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

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

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

$=\sqrt{(8{)}^2}$

$=8units$

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

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

$=\sqrt{16+16}$

$=\sqrt{32}$

$=4\sqrt{2}units$

Similarly, $PR=\sqrt{(4{)}^2+(4{)}^2}$

$=\sqrt{16+16}$

$=\sqrt{32}$

$=4\sqrt{2}units$

Now as pe BPT,

$\Rightarrow$ $\frac{A}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$

$\Rightarrow$ $\frac{4}{8}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{2\sqrt{2}}{4\sqrt{2}}$

$\Rightarrow$ $\frac{1}{2}=\frac{1}{2}=\frac{1}{2}$

$\therefore$ $△ABC\sim △PQR$ ---- Hence proved