Complex Number Problem:Plz help The vertices of a triangle ABC are 1+2j,4-2jand 1-6j.Prove that the triangle is isosceles.?

Distance between (x1,y1) and (x2,y2) is:
SQRT{(x2-x1)^2+(y2-y1)^2]
Coordinates (1,2) and (4,-2) <----------
Distance = SQRT[(1-4)^2)+(2+2)^2]
= sqrt[9+16]
= sqrt[25]
Distance = 5 <----------

Distance between (x1,y1) and (x2,y2) is:
SQRT{(x2-x1)^2+(y2-y1)^2]
Coordinates (4,-2) and (1,-6) <----------
Distance = SQRT[(4-1)^2)+(-2+6)^2]
= sqrt[9+16]
= sqrt[25]
Distance = 5 <---------

Distance between (x1,y1) and (x2,y2) is:
SQRT{(x2-x1)^2+(y2-y1)^2]
Coordinates (1,2) and (1,-6) <---------
Distance = SQRT[(1-1)^2)+(2+6)^2]
= sqrt[0+64]
= sqrt[64]
Distance = 8 <----------

We see that two of its sides are equal to 5 so the triangle is isosceles.

AB = (4 - 2j) - (1 + 2j) = 3 - 4j
BC = (1 - 6j) - (4 - 2J) = -3 - 4j
CA = (1 + 2j) - (1 - 6j) = 8j

Hence,
|AB| = √[3² + (-4)²] = 5
|BC| = √[(-3)² + (-4)²] = 5
|CA| = √[0² + 8²] = 8

Since |AB| = |BC| ≠ |CA|, ABC is an isosceles triangle.

|z-A| = |z-C| means Im(z) = -2 and Im(B) = -2, whilst A+C = 2-4j ≠ 8-4j = 2B, so △ABC is isosceles.

Translation:
The perpendicular bisector of AC passes through B, while B is not the midpoint of AC, so △ABC is isosceles.

(1 + 2 j ) - ( 4 - 2 j ) = - 3 + 4 j
I 3 + 4 j I = 5

( 1 + 2j ) - (1 - 6j) = 8 j
I 8j I = 8

( 4 - 2j ) - (1 - 6j) = 3 + 4 j
I 3 + 4j I = 5

Two sides are equal therefore triangle is isosceles.