What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2? Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?
I started off by noticing that the base of the triangle is 3, so the height must be greater than $\frac{4}{3}$ if the area is greater than 2 because $\frac{1}{2}(3)(h)>2 \rightarrow h >\frac{4}{3}.$
Since $y>\frac{4}{3}, x > \frac{1}{3}$ because the line is $y=x+1$.
So for the area of the triangle to be greater than 2, $P$ can be any point from $(\frac{1}{3}, \frac{4}{3})$ to $(3,4)$ on the line $y=x+1$.
Therefore the probability of the triangle area being greater than 2 is  $\dfrac{\text{length of segment (1/3, 4/3) to (3,4)}}{\text{length of segment (0,1) to (3,4)}}=\dfrac{2/3\sqrt{17}}{3\sqrt{2}}=\dfrac{\sqrt{34}}{9}.$
However the solution says the answer is 8/9, so what am I doing wrong here? I'm also pretty sure I overcomplicated a lot of things, and I was also wondering if there's a simpler solution. Thanks!


The point, $Z$ on the dark green line has coordinates
$x = x(P)$ and $y=area(\triangle PQR)$
So the ratio you are looking for is
$\dfrac{6-2}{6-1.5}=\dfrac{4}{4.5}=\dfrac 89$
