If $a \neq b \neq 0$ , prove that the points $(a, a^{2}), (b, b^{2})$ and $(0, 0)$ will not be collinear.

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Solution

We have,

$A (x_{1}, y_{1}) = (a, a^{2})$

$B (x_{2}, y_{2}) = (b, b^{2})$

$C (x_{3}, y_{3}) = (0, 0)$

We know that,

Three points are not collinear

If, $A r e a o f t r i a n g l e \neq 0$

$[x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0$

$\frac{1}{2} [a (b^{2} - 0) + b (0 - a^{2}) + 0 (b^{2} - a^{2})]$

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

It is not collinear.
Hence proved.

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