A vertical line divides the triangle with vertices $(0,0)$, $(1,1)$, and $(9,1)$ in the $xy\text{-plane}$ into two regions of equal area. The equation of the line is $x=$

$\textbf {(A) } 2.5 \qquad \textbf {(B) } 3.0 \qquad \textbf {(C) } 3.5 \qquad \textbf {(D) } 4.0\qquad \textbf {(E) } 4.5$
1. **Calculate the area of the triangle:**
   The vertices of the triangle are \((0,0)\), \((1,1)\), and \((9,1)\). We can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
   \[
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]
   Substituting the given points:
   \[
   \text{Area} = \frac{1}{2} \left| 0(1 - 1) + 1(1 - 0) + 9(0 - 1) \right| = \frac{1}{2} \left| 0 + 1 - 9 \right| = \frac{1}{2} \left| -8 \right| = 4
   \]

2. **Determine the area of each region:**
   Since the vertical line divides the triangle into two regions of equal area, each region must have an area of:
   \[
   \frac{4}{2} = 2
   \]

3. **Set up the equation for the area of the left region:**
   Let the vertical line be \(x = k\). The left region will be a triangle with vertices \((0,0)\), \((k,1)\), and \((1,1)\). The area of this triangle must be 2. Using the area formula again:
   \[
   \text{Area} = \frac{1}{2} \left| 0(1 - 1) + k(1 - 0) + 1(0 - 1) \right| = \frac{1}{2} \left| 0 + k - 1 \right| = \frac{1}{2} \left| k - 1 \right|
   \]
   Setting this equal to 2:
   \[
   \frac{1}{2} \left| k - 1 \right| = 2 \implies \left| k - 1 \right| = 4
   \]

4. **Solve for \(k\):**
   \[
   k - 1 = 4 \quad \text{or} \quad k - 1 = -4
   \]
   \[
   k = 5 \quad \text{or} \quad k = -3
   \]
   Since \(k\) must be between 0 and 9 (the x-coordinates of the triangle's vertices), we discard \(k = -3\) and keep \(k = 5\).

5. **Verify the solution:**
   The vertical line \(x = 5\) divides the triangle into two regions of equal area, each with an area of 2.

The final answer is \(\boxed{5}\)