Maximizing the area of a triangle with its vertices on a parabola. So, here's the question:
I have the parabola $y=x^2$. Take the points $A=(-1.5, 2.25)$ and $B=(3, 9)$, and connect them with a straight line. Now, I am trying find out how to take a third point on the parabola $C=(x,x^2)$, with $x\in[-1.5,3]$, in such a way that the area of the triangle $ABC$ is maximized.
I have pretty good evidence by trial and error that this point is $(.75, .5625)$ but I have no idea how to prove it. I was trying to work with a gradient, and then Heron's formula, but that was a nightmare to attempt to differentiate. I feel like this is a simple optimization problem but I have no clue how to solve it. Any help is appreciated!
Thanks.

The line through the points $(-1.5,2.25)$ and $(p,p^2)$ is
$$y_1=(p-1.5)x+1.5p,$$
and the line through $(p,p^2)$ and $(3,9)$ is
$$y_2=(p+3)x-3p.$$
(used the basic method here of course, and we may assume that $-1.5<p<3$.)
Now we compute the area of the triangle as the sum of two integrals, each between the line $y=\frac{3}{2}x+4.5$ and our found lines in terms of $p$ over the appropriate bounds.
$$\int_{-1.5}^p (\frac{3}{2}x+4.5-((p-1.5)x+1.5p))\, dx+\int_{p}^3 (\frac{3}{2}x+4.5-((p+3)x-3p))\, dx = -\frac{9}{4}p^2+3.375p+10.125$$
$$\frac{dA}{dp}= 0 =>p = \frac{3.375}{4.5} =0.75$$
$$A_{max} = \frac{729}{64}$$
Thanks
