The three lines `4x - 7y + 10 , x + y = 5 and 7x+ 4y = 15 ` form the sides of a triangle Then the point (1,2) is its

Options

A. centroid
B. incentre
C. orthocentre
D. none of these

Answer

To determine the nature of the point (1,2) with respect to the triangle formed by the lines \(4x - 7y + 10 = 0\), \(x + y = 5\), and \(7x + 4y = 15\), we will follow these steps:

### Step 1: Write the equations of the lines in slope-intercept form
1. **Line 1**: \(4x - 7y + 10 = 0\)
Rearranging gives:
\[7y = 4x + 10\]
\[y = \frac{4}{7}x + \frac{10}{7}\]
Thus, the slope \(m_1 = \frac{4}{7}\).

2. **Line 2**: \(x + y = 5\)
Rearranging gives:
\[y = -x + 5\]
Thus, the slope \(m_2 = -1\).

3. **Line 3**: \(7x + 4y = 15\)
Rearranging gives:
\[4y = -7x + 15\]
\[y = -\frac{7}{4}x + \frac{15}{4}\]
Thus, the slope \(m_3 = -\frac{7}{4}\).

### Step 2: Check if the lines are perpendicular
To check if two lines are perpendicular, we multiply their slopes. If the product is -1, the lines are perpendicular.

- For Line 1 and Line 3:
\[m_1 \cdot m_3 = \frac{4}{7} \cdot -\frac{7}{4} = -1\]
This means Line 1 and Line 3 are perpendicular.

### Step 3: Find the intersection points of the lines
1. **Intersection of Line 1 and Line 3**:
- Set \(4x - 7y + 10 = 0\) and \(7x + 4y - 15 = 0\).
- From Line 1: \(y = \frac{4}{7}x + \frac{10}{7}\).
- Substitute \(y\) into Line 3:
\[7x + 4\left(\frac{4}{7}x + \frac{10}{7}\right) - 15 = 0\]
\[7x + \frac{16}{7}x + \frac{40}{7} - 15 = 0\]
\[49x + 16x + 40 - 105 = 0\]
\[65x - 65 = 0\]
\[x = 1\]
- Substitute \(x = 1\) back into Line 1:
\[4(1) - 7y + 10 = 0\]
\[4 - 7y + 10 = 0\]
\[14 - 7y = 0\]
\[y = 2\]
- Thus, the intersection point is \((1, 2)\).

### Step 4: Determine the nature of the point (1,2)
Since we found that the intersection point of Line 1 and Line 3 is \((1, 2)\), and since Line 1 and Line 3 are perpendicular, the point \((1, 2)\) is the orthocenter of the triangle formed by the three lines.

### Conclusion
The point \((1, 2)\) is the orthocenter of the triangle formed by the lines.

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