The orthocentre of the triangle formed by the lines `x - 7y + 6 = 0, 2x - 5y - 6 = 0 and 7x + y - 8 = 0` is

Options

A. `(8, 2)`
B. `(0,0)`
C. `(1,1)`
D. `(2,8)`

Answer

To find the orthocenter of the triangle formed by the lines \(x - 7y + 6 = 0\), \(2x - 5y - 6 = 0\), and \(7x + y - 8 = 0\), we will follow these steps:

### Step 1: Find the intersection points of the lines
The vertices of the triangle are the intersection points of the given lines.

1. **Intersection of Line 1 and Line 2:**
\[ x - 7y + 6 = 0 \quad \text{(1)} \]
\[ 2x - 5y - 6 = 0 \quad \text{(2)} \]
From (1), we can express \(x\) in terms of \(y\):
\[ x = 7y - 6 \]
Substitute \(x\) into (2):
\[ 2(7y - 6) - 5y - 6 = 0 \]
\[ 14y - 12 - 5y - 6 = 0 \]
\[ 9y - 18 = 0 \implies y = 2 \]
Substitute \(y = 2\) back into (1):
\[ x - 7(2) + 6 = 0 \implies x - 14 + 6 = 0 \implies x = 8 \]
Thus, the intersection point \(A\) is \((8, 2)\).

2. **Intersection of Line 1 and Line 3:**
\[ x - 7y + 6 = 0 \quad \text{(1)} \]
\[ 7x + y - 8 = 0 \quad \text{(3)} \]
From (1), express \(x\) in terms of \(y\):
\[ x = 7y - 6 \]
Substitute into (3):
\[ 7(7y - 6) + y - 8 = 0 \]
\[ 49y - 42 + y - 8 = 0 \]
\[ 50y - 50 = 0 \implies y = 1 \]
Substitute \(y = 1\) back into (1):
\[ x - 7(1) + 6 = 0 \implies x - 7 + 6 = 0 \implies x = 1 \]
Thus, the intersection point \(B\) is \((1, 1)\).

3. **Intersection of Line 2 and Line 3:**
\[ 2x - 5y - 6 = 0 \quad \text{(2)} \]
\[ 7x + y - 8 = 0 \quad \text{(3)} \]
From (2), express \(y\) in terms of \(x\):
\[ y = \frac{2x - 6}{5} \]
Substitute into (3):
\[ 7x + \frac{2x - 6}{5} - 8 = 0 \]
Multiply through by 5 to eliminate the fraction:
\[ 35x + 2x - 6 - 40 = 0 \]
\[ 37x - 46 = 0 \implies x = \frac{46}{37} \]
Substitute \(x = \frac{46}{37}\) back into (2):
\[ 2\left(\frac{46}{37}\right) - 5y - 6 = 0 \]
\[ \frac{92}{37} - 5y - 6 = 0 \]
\[ -5y = 6 - \frac{92}{37} \implies -5y = \frac{222 - 92}{37} = \frac{130}{37} \]
\[ y = -\frac{26}{37} \]
Thus, the intersection point \(C\) is \(\left(\frac{46}{37}, -\frac{26}{37}\right)\).

### Step 2: Determine if the triangle is a right triangle
To check if the triangle is a right triangle, we can find the slopes of the lines:

1. **Slope of Line 1:** \(m_1 = \frac{1}{7}\)
2. **Slope of Line 2:** \(m_2 = \frac{2}{5}\)
3. **Slope of Line 3:** \(m_3 = -7\)

Now, check if any two slopes are negative reciprocals:
- \(m_1 \cdot m_3 = \frac{1}{7} \cdot (-7) = -1\) (indicating that Line 1 and Line 3 are perpendicular).

### Step 3: Find the orthocenter
In a right triangle, the orthocenter is located at the vertex where the right angle is formed. Since Line 1 and Line 3 are perpendicular, the orthocenter is at point \(B(1, 1)\).

### Final Answer
The orthocenter of the triangle formed by the given lines is \((1, 1)\).

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