Show that $3$ distinct points $(p, p^2), (q, q^2)$ and $(r, r^2)$ can never be collinear, using the triangle formula.

Question

Show that $3$ distinct points $(p, p^2), (q, q^2)$ and $(r, r^2)$ can never be collinear, using the triangle formula.

I tried doing it using $$\text{Area }= \frac{1}{2}\big| (x_1y_2 + x_2y_3 + x_3y_1) - (x_2y_1 + x_3y_2 + x_1y_3) \big|$$ and equating it to zero. But I can’t. Plz help.

Answer 1

If factoring area formula seems hard, equivalently, you may row reduce the determinant and show the product along diagonal is never zero: $$\begin{vmatrix}1&p&p^2\\1&q&q^2\\1&r&r^2 \end{vmatrix}\ne 0$$ when $p,q,r$ are distinct.

Answer 2

The area is $$\frac12|pq^2+qr^2+rp^2-qp^2-rq^2-pr^2|=\frac12|(p-q)(q-r)(r-p)|.$$You can guess this factorisation by realising the points will be collinear if any two of $p,\,q,\,r$ are equal, in which case the area will be $0$. But otherwise, the area is clearly non-zero as required.

Answer 3

Use the area formula of the form below $$ A= \frac12 \big| (x_1 - x_3) (y_2 - y_1) - (x_1 - x_2) (y_3 - y_1) \big| $$ $$ =\frac12\big| (p-r)(q^2-p^2)-(p-q)(r^2-p^2)\big|$$ $$=\frac12 \big| (p-r)(q-p)(r-q)\big|\ne 0$$