10. Problem: Let $S$ be the locus of all points $(x, y)$ in the first quadrant such that $\frac{x}{t}+\frac{y}{1-t}=1$ for some $t$ with $0<t<1$. Find the area of $S$.
Solution: Solving for $t$ in the given equation, we get $t^{2}+(y-x-1) t+x=0$. Using the quadratic equation, we get $t=\frac{(x+1-y) \pm \sqrt{(y-x-1)^{2}-4 x}}{2}$. For all valid combinations of $(x, y), t$ is positive and less than 1 (this is easy to see by inspection). All valid combinations of $(x, y)$ are those that make $(y-x-1)^{2}-4 x \geq 0$.

Solving for $y$ in the equation $(y-x-1)^{2}-4 x=0$ yields $y^{2}-(2 x+2) y+(x-1)^{2} \geq 0 \Rightarrow y=(x+1) \pm 2 \sqrt{x}$. In the original equation, it is given that $\frac{x}{t}+\frac{y}{1-t}=1$, and $0<t<1$. This implies that $x, y<1$. Then the only possible $y<1$ that satisfies $(y-x-1)^{2}-4 x=0$ is $y=x+1-2 \sqrt{x}$.

Then to satisfy the inequality $(y-x-1)^{2}-4 x \geq 0$, we must have $y \leq x+1-2 \sqrt{x}$. Recall that this is when $0<y<1$. Hence we integrate in the interval $[0,1]: \in_{0}^{1} x+1-2 \sqrt{x}=\frac{1}{6}$.