Question

If $a\ne b\ne 0$, prove that the points (a, a²), (b, b²) (0, 0) will not be collinear.

Answer 1

Let A(a, a²), B(b, b²) and C(0, 0) be the coordinates of the given points.
We know that the area of triangle having vertices $\left(x_1,y_1\right),\left(x_2,y_2\right)\mathrm{and}\left(x_3,y_3\right)$ 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
$$\begin{gathered} =\left|\frac{1}{2}\left[a\left(b^2-0\right)+b\left(0-a^2\right)+0\left(a^2-b^2\right)\right]\right| \\ =\left|\frac{1}{2}\left(a{b}^2-a^{2}b\right)\right| \\ =\frac{1}{2}\left|ab\left(b-a\right)\right| \\ \ne 0\left(\because a\ne b\ne 0\right) \end{gathered}$$
Since the area of the triangle formed by the points (a, a²), (b, b²) and (0, 0) is not zero, so the given points are not collinear.