If a ≠ b ≠ c, prove that (a, a2), (b, b2), (0, 0) will not be collinear.

Let $\mathrm{A}\left(a, a^{2}\right), \mathrm{B}\left(b, b^{2}\right)$ and $\mathrm{C}(0,0)$ be the coordinates of the given points.

We know that the area of triangle having vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is $\left|\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]\right|$ square units.

So,

Area of ∆ABC

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

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

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

$\neq 0 \quad(\because a \neq b \neq 0)$

Since the area of the triangle formed by the points $\left(a, a^{2}\right),\left(b, b^{2}\right)$ and $(0,0)$ is not zero, so the given points are not collinear.