A) Collinear
B) Vertices of equilateral triangle
C) Vertices of isosceles triangle
D) Vertices of right angled triangle
Correct Answer: A
Solution :
Given that points are \[A(3+4i),B(5-2i)\] and \[C(-1+16i)\] Area of triangle \[=\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=\frac{1}{2}\left| \begin{matrix} 3 & 4 & 1 \\ 5 & -2 & 1 \\ -1 & 16 & 1 \\ \end{matrix} \right|=0\] Hence points are collinear.
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