To find the orthocenter of the triangle formed by the lines \(x + y - 1 = 0\), \(2x + y - 1 = 0\), and \(y = 0\), we will follow these steps:
### Step 1: Find the vertices of the triangle
We need to find the points of intersection of the lines to determine the vertices of the triangle.
1. **Intersection of \(x + y - 1 = 0\) and \(y = 0\)**:
\[
x + 0 - 1 = 0 \implies x = 1 \implies (1, 0)
\]
2. **Intersection of \(2x + y - 1 = 0\) and \(y = 0\)**:
\[
2x + 0 - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2} \implies \left(\frac{1}{2}, 0\right)
\]
3. **Intersection of \(x + y - 1 = 0\) and \(2x + y - 1 = 0\)**:
We can solve these equations simultaneously. From the first equation, we have:
\[
y = 1 - x
\]
Substituting into the second equation:
\[
2x + (1 - x) - 1 = 0 \implies 2x - x + 1 - 1 = 0 \implies x = 0
\]
Now substituting \(x = 0\) back into \(y = 1 - x\):
\[
y = 1 - 0 = 1 \implies (0, 1)
\]
Thus, the vertices of the triangle are \(A(1, 0)\), \(B\left(\frac{1}{2}, 0\right)\), and \(C(0, 1)\).
### Step 2: Find the slopes of the sides of the triangle
1. **Slope of line AB** (between points \(A(1, 0)\) and \(B\left(\frac{1}{2}, 0\right)\)):
\[
\text{slope}_{AB} = \frac{0 - 0}{\frac{1}{2} - 1} = 0
\]
2. **Slope of line AC** (between points \(A(1, 0)\) and \(C(0, 1)\)):
\[
\text{slope}_{AC} = \frac{1 - 0}{0 - 1} = -1
\]
3. **Slope of line BC** (between points \(B\left(\frac{1}{2}, 0\right)\) and \(C(0, 1)\)):
\[
\text{slope}_{BC} = \frac{1 - 0}{0 - \frac{1}{2}} = -2
\]
### Step 3: Find the angles at each vertex
Using the slopes, we can find the angles:
1. **Angle at A**:
\[
\tan A = \left| \frac{\text{slope}_{AC} - \text{slope}_{AB}}{1 + \text{slope}_{AC} \cdot \text{slope}_{AB}} \right| = \left| \frac{-1 - 0}{1 + 0} \right| = 1 \implies A = 45^\circ
\]
2. **Angle at B**:
\[
\tan B = \left| \frac{\text{slope}_{BC} - \text{slope}_{AB}}{1 + \text{slope}_{BC} \cdot \text{slope}_{AB}} \right| = \left| \frac{-2 - 0}{1 + 0} \right| = 2 \implies B = 63.43^\circ
\]
3. **Angle at C**:
\[
\tan C = \left| \frac{\text{slope}_{AC} - \text{slope}_{BC}}{1 + \text{slope}_{AC} \cdot \text{slope}_{BC}} \right| = \left| \frac{-1 - (-2)}{1 + (-1)(-2)} \right| = \frac{1}{3} \implies C = 18.43^\circ
\]
### Step 4: Calculate the orthocenter coordinates
Using the formula for the orthocenter \((h, k)\):
\[
h = \frac{x_1 \tan A + x_2 \tan B + x_3 \tan C}{\tan A + \tan B + \tan C}
\]
\[
k = \frac{y_1 \tan A + y_2 \tan B + y_3 \tan C}{\tan A + \tan B + \tan C}
\]
Substituting the values:
- \(A(1, 0)\), \(B\left(\frac{1}{2}, 0\right)\), \(C(0, 1)\)
- \(\tan A = 1\), \(\tan B = 2\), \(\tan C = \frac{1}{3}\)
Calculating \(h\):
\[
h = \frac{1 \cdot 1 + \frac{1}{2} \cdot 2 + 0 \cdot \frac{1}{3}}{1 + 2 + \frac{1}{3}} = \frac{1 + 1 + 0}{3 + \frac{1}{3}} = \frac{2}{\frac{10}{3}} = \frac{6}{10} = \frac{3}{5}
\]
Calculating \(k\):
\[
k = \frac{0 \cdot 1 + 0 \cdot 2 + 1 \cdot \frac{1}{3}}{1 + 2 + \frac{1}{3}} = \frac{0 + 0 + \frac{1}{3}}{3 + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{10}{3}} = \frac{1}{10}
\]
### Step 5: Calculate \(\frac{1}{k^2}\)
Now we need to find \(\frac{1}{k^2}\):
\[
k = \frac{1}{10} \implies k^2 = \left(\frac{1}{10}\right)^2 = \frac{1}{100}
\]
\[
\frac{1}{k^2} = 100
\]
### Final Answer
Thus, the value of \(\frac{1}{k^2}\) is \(\boxed{100}\).
