Army Group X Practice Set 1

Let P(4, 4) , Q (3,5) , R( - 1, - 1 ) be the vertices of ∆PQR.
Then, |PQ|=√(3−4)2+(5−4)2
=√1+1=√2
[∵distance‌between‌two‌points(x1,y1)
and (x2,y2)=√(x2−x1)2+(y2−y1)2
|QR|=√(−1−3)2+(−1−5)2
=√16+36=√52
|RP|=√(4+1)2+(4+1)2
=√25+25=√50=5√2
Thus, QR2=PQ2+PR2
[∵52=2+50]
∴ ∠ QRP is a right angle [by Pythagorastheorem ]
Hence ,∆PQR is a right angle triangle.