Question

Show that point P (-2,2) Q (2,2) R (2,7) are vertices of a right angle triangle

Answer 1

since they are Pythagorean triplets...
I hope it helps...

Answer 2

Therefore the points P (-2,2) Q (2,2) R (2,7) are vertices of a right angle triangle

Hence proved.

Step-by-step explanation:

Given:

Here we have to prove that points P (-2,2) Q (2,2) R (2,7) are vertices of a right angle triangle.

In a right angles triangle PQR, right angled at Q, according to the pythagoras theorem.

$PQ^2+QR^2=PR^2$

According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by

$d=\sqrt{(a-c)^2+(b-d)^2}$                  (1)

For the given points Distance between P and Q is

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

$PQ=\sqrt{16}$

$QR=\sqrt{(2-2)^2+(7-2)^2}$

$QR=\sqrt{25}$

$PR=\sqrt{(-2-2)^2+(2-7)^2}$

$PR=\sqrt{16+25}$

$PR=\sqrt{41}$

Therrefore,

$PQ^{2} =16$            

$QR^2=25$

$PR^2=41$                     (2)

PQ^2+OR^2=16+25=41            (3)

From equation 2 and 3,

$PQ^2+QR^2=PR^2$.

Hence proved.