Question

The coordinates of vertices of triangle ABC are A(0, 0), B(0, 2) and, C(2, 0). Prove that triangle ABC is an isosceles triangle.

Answer 1

first
let
A(0, 0),=(X1,y1)
B(0, 2)=(X2,y2)

distance AB =ROOTS (X2-X1)^2+(Y2-Y1)^2

=root(0-0)^2+(2-0)^2
=root4
2unit

similar B.C.
B(0, 2) =(X1,y1)
C(2, 0) =(X2,y2)

distance BC=ROOTS (X2-X1)^2+(Y2-Y1)^2
=root(2-0)^2+(0-2)^2
=ROOT 4+4
=ROOT 8
=2root2 unit


A(0, 0),=(X1,y1)
C(2, 0)=(X2,y2)


distance AC=ROOTS (X2-X1)^2+(Y2-Y1)^2

=ROOT (2-0)^2+(0-0)^2

=ROOT 4
=2 unit

hence AB=AC SO,
ABC IS AN ISOSCELES TRIANGLE