Question

Verify that points $P(-2,2),Q(2,2)$ and $R(2,7)$ are vertices of right angled triangle.

Answer 1

Given,

$P(-2,2)$

$Q(2,2)$

$R(2,7)$

$\Rightarrow \left|PQ\right|=\sqrt{{(-2-2)}^2+{(2-2)}^2}=\sqrt{{\left(4\right)}^2}=4units$

$\Rightarrow \left|PR\right|=\sqrt{{(-2-2)}^2+{(2-7)}^2}=\sqrt{{\left(4\right)}^2+5^2}=\sqrt{16+25}=\sqrt{41}units$

$\Rightarrow \left|PQ\right|=\sqrt{{(2-2)}^2+{(2-7)}^2}=\sqrt{{\left(5\right)}^2}=5units$

we can clearly see that

$\Rightarrow {\left|PR\right|}^2={\left|PQ\right|}^2+{\left|QR\right|}^2$ ( Pythagoras theorem )

since ${\left|PQ\right|}^2=16$

and ${\left|QR\right|}^2=25$

$\therefore {\left|PQ\right|}^2+{\left|QR\right|}^2=16+25=41$

and we have $\left|PR\right|=\sqrt{41}$

$\therefore {\left|PR\right|}^2=41units$

Hence, we have 'PR' as the hypotenuse and PQ and QR are the other two legs of the right angled triangle.