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!
 A: The area of the triangle is given by the function $\displaystyle A(X) = \frac{3}{2}(X+1)$, where X is uniformly distributed in $[0,3]$, so
\begin{equation}
\mathbb{P}(A(X) \geq 2) = \mathbb{P}\left(X\geq\frac{1}{3} \right) = \frac{1}{3} \left(3 - \frac{1}{3} \right) = \frac{8}{9}
\end{equation}
A: 
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
$\dfrac{6-2}{6-1.5}=\dfrac{4}{4.5}=\dfrac 89$
A: If the upper vertex of the triangle is $(x,y)$, the area of the triangle formed by $(0,0)$, $(3,0)$, and $(x,y)$ is ${1 \over 2}bh$ or ${3h \over 2}$.
So the probability the area is at least $2$ is the probability that ${3h \over 2} > 2$, or equivalently the probability that $h > {4 \over 3}$. Since $h$ is uniformly distributed in $[1,4]$, the probability of this event is ${4 - {4 \over 3} \over 4 - 1} = {8 \over 9}$.
A: The base is of length $3;$ the height is $h = 1 + U,$ where $U \sim \mathsf{Unif}(1,4).$ So the following R code finds areas of a million triangles and finds the proportion with areas exceeding 2.
With a million such simulated triangles the desired probability
$P(A > 2)$ should be estimated to two decimal places.
set.seed(2021)
a = .5*3*runif(10^6, 1, 4)
mean(a > 2)
[1] 0.889393   # aprx P(A > 2)
8/9
[1] 0.8888889  # exact

Congratulations on asking a popular questions. Lots of answers are appearing almost simultaneously. My simulation
is essentially the same idea as in @W's Answer (+1)---and also $Zarrax' (+1).
