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 (x₁, y₁), (x₂, y₂) and (x₃, y₃) is ∣∣`1/2`[x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)]∣∣ square units.

So,

Area of ∆ABC

`= |1/2|a(b^2 - 0) + b(0 - a^2) + 0(a^2 -  b^2)||`

`=|1/2(ab^2 - a^2b)|`

`= 1/2|ab(b -a)|`

`!= 0     (∵ a!= b != 0)`

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.