Question

prove that (2, -2), (-2, 1) and (5, 2) are vertices of a right angled triangle

Answer 1

let A (2,-2) B (-2,1) and c (5,2)

now ,
use distance formula ,
AB = {(2+2)^2 + (-2-1)^2}^1/2= 5
BC ={(5+2)^2 +(2-1)^2}^1/2 = (50)^1/2
CA = {(5-2)^2 +(2 + 2)^2 }^1/2 = 5
now ,
AB^2 =25
BC^2 =50
CA^2 = 25
you can see ,
AB^2 +CA^2 =BC^2
according to Pythagoras theorem ,
ABC is right angle triangle .

Answer 2

ΔABC:     A(2,-2),   B(-2, 1) ,  C(5,2)

Slope of AB = (1+2)/(-2-2) = -3/4
Slope of BC = (2-1)/(5-(-2)) = 1/7
Slope of CA = (2-(-2)) / (5-2) = 4/3

Product of slopes of AB & CA = -1.
    Clearly  AB ⊥ CA .  So ΔABC is right angled at A.