Area Of Triangle in Coordinate Geometry

To find the length $L$ of the perpendicular from the origin to the line $y=\frac{-3}{4}x-\frac{15}{4}$ , let $\begin{aligned} y&=\dfrac 43x=\dfrac{-3}{4}x-\dfrac{15}{4}\\ \Longrightarrow (x,y)&=\left(\dfrac{-9}{5},\dfrac{-12}{5}\right) \end{aligned}$ so that $L=\sqrt{\left(\dfrac{-9}{5}\right)^2+\left(\dfrac{-12}{5}\right)^2}=3.$ Then the area of the triangle is $3\times \sqrt{9^2-3^2}=18\sqrt{2}=\boxed{\sqrt{648}}.$