# Solve the following

Question:

If $a \neq b \neq c$, prove that the points $\left(a, a^{2}\right),\left(b, b^{2}\right),\left(c, c^{2}\right)$ can never be collinear.

Solution:

GIVEN: If $a \neq b \neq c$

TO PROVE: that the points $\left(a, a^{2}\right),\left(b, b^{2}\right),\left(c, c^{2}\right)$, can never be collinear.

PROOF:

We know three points $\left(x_{1}, y_{1}\right),\left(x_{2} y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ are collinear when $\frac{1}{2} \| x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$

Now taking three points $\left(a, a^{2}\right),\left(b, b^{2}\right),\left(c, c^{2}\right)$,

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

$=\frac{1}{2}\left|a b^{2}-a c^{2}+b c^{2}-b a^{2}+c a^{2}-c b^{2}\right|$

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

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

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

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

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

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

Also it is given that

$a \neq b \neq c$

Hence area of triangle made by these points is never zero. Hence given points are never collinear.