P(-2, 5), Q(-1 , 1), R(7,3)
Slope of line PQ = (5 - 1)/(-2 + 1) = -4
Slope of line QR = (3 - 1)/(7 + 1) = 2/8 = 1/4
Perpendicular lines, slope is opposite and reciprocal
So line PQ is perpendicular with line QR
Conclusion: Triangle PQR is a right triangle at Q
Or you can solve by find the distance of each side
Distance of PQ = √[(-1 +2)^2 + (1 - 5)^2] = √17
Distance of QR = √[(7+1)^2 + (3-1)^2] = √(64+4) = √68
Distance of PR = √[(7 +2)^2 + (3 - 5)^2] = √(81 + 4) = √85
Pythagorean theorem: If a right triangle then c^2 = a^2 + b^2
(√85)^2 = (√17)^2 + (√68)^2
85 = 17 + 68
85 = 85
Conclusion: Triangle PQR is a right triangle