Question

proove that point A(2,-2),B(14,10)andC(11,13)are vertices of right angle triangle

Answer 1

we should first find the square of distance
(AB)^2 = (2-14)^2 + (-2-10)^2 = 144 + 144 =288
(BC)^2 = (14-11)^2 + (10-13)^2 = 9 + 9 = 18
( AC)^2 = ( 11-2)^2+ (13+2) ^2= 81 + 225 = 306

which implies (AB)^2 + (BC)^2 = AC)^2
so they are the vertices of a right triangle

Answer 2

A (2,-2)
Hyp=√h2+b2
√(2)²+(-2)²
√4+4
√8

B (14+10)
Hyp=√h²+B²
√(14)²+(10)²
√196+100
√296

C (11,13)
Hyp=√h²+b²
√(11)²+(13)²
√121+16