Prove that $A (1, 1), B (2, 2)$ and $C (3, 3)$ are collinear.

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Solution

$\begin{matrix} W e h a v e t h e g i v e p o i n t s a r e (1, 1) (2, 2) (3, 3) N o w, A r e a o f Δ = \frac{1}{2} [x_{1} (y^{2} - y^{3}) + x_{2} (y^{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [1 (2 - 3) + 2 (3 - 1) + 3 (1 - 2)] = \frac{1}{2} [1 (- 1) + 2 (2) + 3 (- 1)] = \frac{1}{2} [- 1 + 4 - 3] = \frac{1}{2} [4 - 4] = 0 H e n c e, t h e a r e a o f t r i a n g l e a r e 0 s o, t h e g i e n p o i t s a r e c o l i n e a r . \end{matrix}$

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