A regression line in a corner

Idea. The least-squares line estimates the best linear predictor of $y$ from $x$. If the conditional mean of $y$ given $x$ is already linear, it is that predictor, so there is nothing to compute beyond the conditional mean.

Step 1: the conditional distribution. For $x\in [\frac{1}{2},1]$, $y$ ranges over $[\frac{3}{2}-x,1]$, and since the points are uniform on the triangle, $y$ is uniform on that interval given $x$.

Step 2: the conditional mean. $$\mathbb{E}[y\mid x]=\frac{(\frac{3}{2}-x)+1}{2}=\frac{5}{4}-\frac{x}{2}.$$

Step 3. This is linear in $x$, so it is the population regression line: slope $k=-\frac{1}{2}$, intercept $b=\frac{5}{4}$.

Check by moments. $x$ has density $8(x-\frac{1}{2})$ on $[\frac{1}{2},1]$, so $\mathbb{E}[x]=\mathbb{E}[y]=\frac{5}{6}$, $\mathrm{Var}(x)=\frac{1}{72}$ and $\mathrm{Cov}(x,y)=-\frac{1}{144}$. Then $k=\frac{-1/144}{1/72}=-\frac{1}{2}$ and $b=\frac{5}{6}+\frac{1}{2}\cdot \frac{5}{6}=\frac{5}{4}$.

The trap. The triangle's hypotenuse has slope $-1$, and it is tempting to answer with the line $x+y=\frac{3}{2}$ or a parallel one. The regression line is flatter, because it averages $y$ vertically.