Find the area of the triangle under certain preconditions 
With vertices $(0, 0)$, $(b, a)$, $(x, y)$, prove the area of this triangle is $\frac{|by - ax|}{2}$.

We know area of a triangle = $\frac{rh}{2}$. ($r$ is the base.) Well, we have $r =$ the distance from the origin to $(x, y)$ and the height is the line perpendicular to $y = mx$, which is $y = -\frac{x}{m} + b'$, intersecting the points $(x_0, y_0)$ on the line $y = mx$ and $(b, a)$.
So then, the altitude from my calculations was $ h =\sqrt{1+\frac{1}{m^2}}|x_0 - b|$ The base is just the distance to the origin from $(x, y)$, which is $d = \sqrt{x^2+y^2}$. But I don't see how this equals what I'm trying to prove.

Perhaps you'll find it easier to compute the area by expressing it as half the magnitude of the vector product $\left\langle b,a,0\right\rangle \times \left\langle x,y,0\right\rangle$. Indeed, if $\theta$ is the interior angle between the base and the side $\ell$ from the origin to $(x,y)$, then $h = \ell\sin \theta$. So the area is $(1/2)r\ell\sin \theta$. This is the same as $(1/2)\|\left\langle b,a,0\right\rangle \times \left\langle x,y,0\right\rangle\|$. The cross product equals $\left\langle 0, 0, by - ax \right\rangle$. Therefore, the area is $(1/2)\|\left\langle 0, 0, by - ax\right\rangle\| = (1/2)|by - ax|$.
