The orthocentre of triangle formed by lines `x+y-1=0, 2x+y-1=0 and y=0` is (h, k), then `(1)/(k^(2))=`

Answer

To solve the problem of finding the orthocenter of the triangle formed by the lines \(x + y - 1 = 0\), \(2x + y - 1 = 0\), and \(y = 0\), and then calculating \(\frac{1}{k^2}\) where the orthocenter is \((h, k)\), we can follow these steps:

### Step 1: Find the Points of Intersection
First, we need to find the vertices of the triangle by determining the points of intersection of the lines.

1. **Intersection of \(x + y - 1 = 0\) and \(y = 0\)**:
\[ x + 0 - 1 = 0 \implies x = 1 \implies (1, 0) \]
This gives us point \(B(1, 0)\).

2. **Intersection of \(2x + y - 1 = 0\) and \(y = 0\)**:
\[ 2x + 0 - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2} \implies \left(\frac{1}{2}, 0\right) \]
This gives us point \(C\left(\frac{1}{2}, 0\right)\).

3. **Intersection of \(x + y - 1 = 0\) and \(2x + y - 1 = 0\)**:
- From \(x + y = 1\) (1)
- From \(2x + y = 1\) (2)

Subtract (1) from (2):
\[ (2x + y) - (x + y) = 1 - 1 \implies x = 0 \]
Now substitute \(x = 0\) back into (1):
\[ 0 + y = 1 \implies y = 1 \implies (0, 1) \]
This gives us point \(A(0, 1)\).

### Step 2: Determine the Altitudes
Next, we need to find the equations of the altitudes of the triangle.

1. **Altitude from \(B(1, 0)\) to side \(AC\)**:
- The slope of line \(AC\) (from \(A(0, 1)\) to \(C\left(\frac{1}{2}, 0\right)\)):
\[ \text{slope of } AC = \frac{0 - 1}{\frac{1}{2} - 0} = -2 \]
- The slope of the altitude \(BE\) is the negative reciprocal:
\[ \text{slope of } BE = \frac{1}{2} \]
- Using point-slope form for the altitude through \(B(1, 0)\):
\[ y - 0 = \frac{1}{2}(x - 1) \implies y = \frac{1}{2}x - \frac{1}{2} \]

2. **Altitude from \(C\left(\frac{1}{2}, 0\right)\) to side \(AB\)**:
- The slope of line \(AB\):
\[ \text{slope of } AB = \frac{1 - 0}{0 - 1} = -1 \]
- The slope of the altitude \(CD\) is the negative reciprocal:
\[ \text{slope of } CD = 1 \]
- Using point-slope form for the altitude through \(C\left(\frac{1}{2}, 0\right)\):
\[ y - 0 = 1\left(x - \frac{1}{2}\right) \implies y = x - \frac{1}{2} \]

### Step 3: Find the Orthocenter
Now, we find the intersection of the two altitude lines:

1. **Equate the two altitude equations**:
\[ \frac{1}{2}x - \frac{1}{2} = x - \frac{1}{2} \]
- Solving for \(x\):
\[ \frac{1}{2}x = x \implies 0 = \frac{1}{2}x \implies x = 0 \]
- Substitute \(x = 0\) into one of the altitude equations to find \(y\):
\[ y = \frac{1}{2}(0) - \frac{1}{2} = -\frac{1}{2} \]
Thus, the orthocenter is \((0, -\frac{1}{2})\).

### Step 4: Calculate \(\frac{1}{k^2}\)
Here, \(k = -\frac{1}{2}\):
\[ k^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \]
Thus,
\[ \frac{1}{k^2} = \frac{1}{\frac{1}{4}} = 4 \]

### Final Answer
The value of \(\frac{1}{k^2}\) is \(4\).