If $a ≠ b ≠ 0$, prove that the points $(a, a^2), (b, b^2), (0, 0)$ are never collinear.

Given:

Given points are $(a, a^2), (b, b^2), (0, 0)$.

$a ≠ b ≠ 0$

To do:

We have to prove that the given points are never collinear.

Solution:

Let $A(a, a^2), B(b, b^2)$ and $C(0, 0)$ be the vertices of $\triangle ABC$.

We know that,

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, 

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( ABC=\frac{1}{2}[a(b^2-0)+b(0-a^2)+0(a^2-b^2)] \)

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

\( =\frac{1}{2}[ab(b-a)] \)

\( ≠0 \)       (Since $a ≠ b ≠ 0$)

Here,

The area of $\triangle ABC$ is not equal to zero.

Therefore, points $A, B$ and $C$ are not collinear.

Hence proved.  

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Updated on: 10-Oct-2022

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