If $a \neq b \neq 0,$ prove that the points $(a, a^{2}), (b, b^{2}), (0, 0)$ are never collinear.

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Solution

Area of triangle = $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | a (b^{2} - 0) + b (0 - a^{2}) + 0 (a^{2} - b^{2}) |$

$= \frac{1}{2} | a b^{2} - b a^{2} = 0 |$

$= \frac{1}{2} | a b (b - a) |$

Since $a \neq b \neq 0$ , the area can't be zero. Hence, the points can't be collinear.

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