# If $a ‚Č† b ‚Č† c$, prove that the points $(a, a^2), (b, b^2), (c, c^2)$ can never be collinear.

Given:

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

$a ≠ b ≠ c$

To do:

We have to prove that the given points can never be collinear.

Solution:

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

We know that,

If the points $A, B$ and $C$ are collinear then the area of $\triangle ABC$ is zero.

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

$=\frac{1}{2}[ab^2-ac^2+bc^2-a^2b+ca^2-b^2c]$

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

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

$=\frac{1}{2}(b-c)(a-b)(c-a)$
$=\frac{1}{2}(a-b)(b-c)(c-a)$     (Since $a ≠ b ≠ c$)

Here,

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

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

Hence proved.‚Ää‚Ää‚Ää

Updated on: 10-Oct-2022

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