Question

If $a\ne b\ne c$, prove that the points $(a,a^2),(b,b^2),(c,c^2)$ can never be collinear.

Answer 1

We know that the area of a triangle 0 when $a=b=c,$

Let Δ be the area of the triangle formed by the points

$\Delta =\frac{1}{2}\mid y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\mid$

Here, $x_1=a,y_1=a^2,x_2=b,y_2=b^2,x_3=c,y_3=c^2$

$\therefore \Delta =\frac{1}{2}\left|\right(a{b}^2+b{c}^2+c{a}^2)-(a^{2}b+b^{2}c+c^{2}a\left)\right|\Delta =\frac{1}{2}\mid (a^{2}c-a^{2}b)+(a{b}^2-a{c}^2)+(b{c}^2-b^{2}c)\mid \Delta =\frac{1}{2}\mid -a^2(b-c)+a(b^2-c^2)-bc(b-c)\mid \Delta =\frac{1}{2}\left|(b-c)\{-a^2+a(b+c)+bc\}\right|\Delta =\frac{1}{2}\left|(b-c)\{-a^2+ab+ac+bc\}\right|$

Hence, it is given that the area of triangle 0 when $a=b=c,$ but $a\ne b\ne c$ so the area of triangle can't be zero. That's why all the given three points can never be collinear.