If A, B, C are represented by 3 + 4i, 5 - 2i , -1 + 16i, then A, B, C are
[RPET 1986]

Collinear

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Solution

The correct option is A Collinear
Given that points are A(3 + 4i), B(5 - 2i) and C(-1 + 16i)
Area of triangle = $\frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 3 & 4 & 1 5 & - 2 & 1 - 1 & 16 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Hence points are collinear.

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If A, B, C are represented by 3 + 4i, 5 - 2i , -1 + 16i, then A, B, C are [RPET 1986]

The correct option is A Collinear Given that points are A(3 + 4i), B(5 - 2i) and C(-1 + 16i) Area of triangle = 12∣∣ ∣∣x1y11x2y21x3y31∣∣ ∣∣=12∣∣ ∣∣3415−21−1161∣∣ ∣∣=0 Hence points are collinear.

If A, B, C are represented by 3 + 4i, 5 - 2i , -1 + 16i, then A, B, C are
[RPET 1986]