The graph of the equation $x^y =y^x$ in the first quadrant consists of a straight line and a curve. Find the coordinates of the intersection of the line and the curve.
1. **Rewrite the given equation:**
   \[
   x^y = y^x
   \]
   Taking the natural logarithm on both sides, we get:
   \[
   \log(x^y) = \log(y^x)
   \]
   Using the logarithm power rule, this simplifies to:
   \[
   y \log x = x \log y
   \]
   Dividing both sides by \(xy\), we obtain:
   \[
   \frac{\log x}{x} = \frac{\log y}{y}
   \]

2. **Define a function \(f(x)\):**
   Let:
   \[
   f(x) = \frac{\log x}{x}
   \]
   We need to find the critical points of \(f(x)\) to understand its behavior. Compute the derivative \(f'(x)\):
   \[
   f'(x) = \frac{d}{dx} \left( \frac{\log x}{x} \right)
   \]
   Using the quotient rule:
   \[
   f'(x) = \frac{(1/x) \cdot x - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}
   \]

3. **Analyze the critical points:**
   Set \(f'(x) = 0\):
   \[
   \frac{1 - \log x}{x^2} = 0 \implies 1 - \log x = 0 \implies \log x = 1 \implies x = e
   \]
   To determine the nature of this critical point, consider the second derivative \(f''(x)\):
   \[
   f''(x) = \frac{d}{dx} \left( \frac{1 - \log x}{x^2} \right)
   \]
   Using the quotient rule again:
   \[
   f''(x) = \frac{(0 - \frac{1}{x}) \cdot x^2 - (1 - \log x) \cdot 2x}{x^4} = \frac{-x - 2x(1 - \log x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{-3x + 2x \log x}{x^4} = \frac{2x \log x - 3x}{x^4} = \frac{2 \log x - 3}{x^3}
   \]
   Evaluate \(f''(x)\) at \(x = e\):
   \[
   f''(e) = \frac{2 \log e - 3}{e^3} = \frac{2 \cdot 1 - 3}{e^3} = \frac{-1}{e^3} < 0
   \]
   Since \(f''(e) < 0\), \(x = e\) is a local maximum.

4. **Behavior of \(f(x)\):**
   - \(f(x)\) increases on \((0, e]\)
   - \(f(x)\) decreases on \([e, \infty)\)

5. **Intersection of the line and the curve:**
   The line \(x = y\) intersects the curve \(f(x) = f(y)\) at the point where \(x = y\). Since \(f(x)\) has a maximum at \(x = e\), the intersection point in the first quadrant is:
   \[
   (x, y) = (e, e)
   \]

The final answer is \(\boxed{(e, e)}\)