A point $(x,y)$ in the first quadrant lies on a line with intercepts $(a,0)$ and $(0,b)$, with $a,b > 0$. Rectangle $M$ has vertices $(0,0)$, $(x,0)$, $(x,y)$, and $(0,y)$, while rectangle $N$ has vertices $(x,y)$, $(x,b)$, $(a,b)$, and $(a,y)$. What is the ratio of the area of $M$ to that of $N$?

[i]Proposed by Eugene Chen[/i]
1. **Determine the equation of the line:**
   The line passes through the intercepts \((a,0)\) and \((0,b)\). The equation of a line in standard form given intercepts \(a\) and \(b\) is:
   \[
   \frac{x}{a} + \frac{y}{b} = 1
   \]
   Multiplying through by \(ab\) to clear the denominators, we get:
   \[
   bx + ay = ab
   \]

2. **Calculate the area of rectangle \(M\):**
   Rectangle \(M\) has vertices \((0,0)\), \((x,0)\), \((x,y)\), and \((0,y)\). The area of rectangle \(M\) is:
   \[
   \text{Area of } M = x \cdot y = xy
   \]

3. **Calculate the area of rectangle \(N\):**
   Rectangle \(N\) has vertices \((x,y)\), \((x,b)\), \((a,b)\), and \((a,y)\). The area of rectangle \(N\) is:
   \[
   \text{Area of } N = (a - x)(b - y)
   \]
   Expanding this expression, we get:
   \[
   \text{Area of } N = ab - ay - bx + xy
   \]

4. **Simplify the area of rectangle \(N\):**
   From the equation of the line \(bx + ay = ab\), we can substitute \(ab\) with \(bx + ay\):
   \[
   \text{Area of } N = ab - ay - bx + xy = (bx + ay) - ay - bx + xy
   \]
   Simplifying further, we see that the terms \(bx\) and \(ay\) cancel out:
   \[
   \text{Area of } N = xy
   \]

5. **Determine the ratio of the areas:**
   The ratio of the area of \(M\) to the area of \(N\) is:
   \[
   \frac{\text{Area of } M}{\text{Area of } N} = \frac{xy}{xy} = 1
   \]

Conclusion:
\[
\boxed{1}
\]