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.

hint: Let $M=(x,x^2)$ be the point on the parabola $y = x^2$, find the distance from this point to the base $AB$ where $A = (-1.5,2.25), B = (3,9)$. This distance is easy to find and it is a function of $x$. You can use calculus to find the max distance, and this corresponds to the max area since the base is fixed.
