Answer: \(PQ^2+QR^2=PR^2\), so ΔPQR is right angled at Q.
\[PQ^2=(2+2)^2+(2-2)^2=16\]
\[QR^2=(2-2)^2+(7-2)^2=25\]
\[PR^2=(2+2)^2+(7-2)^2=41\]
\(PQ^2+QR^2=16+25=41=PR^2\). By the converse of Pythagoras theorem, \(\angle PQR=90^\circ\), so P, Q, R are vertices of a right angled triangle.