The co-ordinates of the orthocentre of the triangle bounded by the lines, `4x -7y + 10=0;x+y=5` and `7x + 4y=15` is

Options

A. (-1,-2)
B. (1,-2)
C. (-1,2)
D. (1,2)

Answer

To find the coordinates of the orthocenter of the triangle formed by the lines \(4x - 7y + 10 = 0\), \(x + y = 5\), and \(7x + 4y = 15\), we will follow these steps:

### Step 1: Find the intersection points of the lines
We need to find the vertices of the triangle formed by the intersection of the three lines.

**Intersection of Line 1 and Line 2:**
1. The equations are:
- Line 1: \(4x - 7y + 10 = 0\)
- Line 2: \(x + y = 5\)

We can express \(y\) from Line 2:
\[ y = 5 - x \]

Substituting \(y\) in Line 1:
\[ 4x - 7(5 - x) + 10 = 0 \]
\[ 4x - 35 + 7x + 10 = 0 \]
\[ 11x - 25 = 0 \]
\[ 11x = 25 \implies x = \frac{25}{11} \]

Now substituting \(x\) back to find \(y\):
\[ y = 5 - \frac{25}{11} = \frac{55 - 25}{11} = \frac{30}{11} \]

So, the intersection point \(A\) is:
\[ A\left(\frac{25}{11}, \frac{30}{11}\right) \]

**Intersection of Line 1 and Line 3:**
2. The equations are:
- Line 1: \(4x - 7y + 10 = 0\)
- Line 3: \(7x + 4y = 15\)

We can express \(y\) from Line 1:
\[ 7y = 4x + 10 \implies y = \frac{4x + 10}{7} \]

Substituting \(y\) in Line 3:
\[ 7x + 4\left(\frac{4x + 10}{7}\right) = 15 \]
\[ 7x + \frac{16x + 40}{7} = 15 \]
Multiplying through by 7 to eliminate the fraction:
\[ 49x + 16x + 40 = 105 \]
\[ 65x = 65 \implies x = 1 \]

Now substituting \(x\) back to find \(y\):
\[ 4(1) - 7y + 10 = 0 \implies 4 - 7y + 10 = 0 \implies -7y = -14 \implies y = 2 \]

So, the intersection point \(B\) is:
\[ B(1, 2) \]

**Intersection of Line 2 and Line 3:**
3. The equations are:
- Line 2: \(x + y = 5\)
- Line 3: \(7x + 4y = 15\)

From Line 2, express \(y\):
\[ y = 5 - x \]

Substituting \(y\) in Line 3:
\[ 7x + 4(5 - x) = 15 \]
\[ 7x + 20 - 4x = 15 \]
\[ 3x = -5 \implies x = -\frac{5}{3} \]

Now substituting \(x\) back to find \(y\):
\[ y = 5 - \left(-\frac{5}{3}\right) = 5 + \frac{5}{3} = \frac{15 + 5}{3} = \frac{20}{3} \]

So, the intersection point \(C\) is:
\[ C\left(-\frac{5}{3}, \frac{20}{3}\right) \]

### Step 2: Verify if the triangle is a right triangle
To find the orthocenter, we need to check if the triangle is a right triangle. We can do this by checking if any two lines are perpendicular.

The slopes of the lines are:
- For Line 1: \(4x - 7y + 10 = 0 \implies y = \frac{4}{7}x + \frac{10}{7}\) (slope = \(\frac{4}{7}\))
- For Line 2: \(x + y = 5 \implies y = -x + 5\) (slope = \(-1\))
- For Line 3: \(7x + 4y = 15 \implies y = -\frac{7}{4}x + \frac{15}{4}\) (slope = \(-\frac{7}{4}\))

Now, check if any two slopes are negative reciprocals:
- Slope of Line 1 and Line 2: \(\frac{4}{7} \cdot (-1) = -\frac{4}{7}\) (not perpendicular)
- Slope of Line 1 and Line 3: \(\frac{4}{7} \cdot \left(-\frac{7}{4}\right) = -1\) (perpendicular)

Since Line 1 and Line 3 are perpendicular, the triangle is a right triangle.

### Step 3: Find the orthocenter
In a right triangle, the orthocenter is located at the vertex where the right angle is formed. Therefore, the orthocenter is at point \(B(1, 2)\).

### Final Answer
The coordinates of the orthocenter of the triangle are:
\[ (1, 2) \]