The mid points of the sides of a triangle are (5,0),(5,12) and (0,12), then orthocentre of this triangle is?

A.(0,0)
B.(0,24)
C.(10,0)
D.None of these

We are given the midpoints of the sides of a triangle: (5, 0), (5, 12), and (0, 12). Let's first find the coordinates of the triangle's vertices.

### Step 1: Find the vertices of the triangle
The midpoints of the sides of a triangle divide each side into two equal parts. Using the midpoint formula, we can deduce the vertices.

Let the vertices of the triangle be \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \).

- Midpoint of \( AB \) is \( (5, 0) \)
- Midpoint of \( BC \) is \( (5, 12) \)
- Midpoint of \( CA \) is \( (0, 12) \)

The midpoint formula is:

\[
\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

From the midpoint of \( AB \), we have:

\[
\frac{x_1 + x_2}{2} = 5 \quad \text{and} \quad \frac{y_1 + y_2}{2} = 0
\]

This gives:

\[
x_1 + x_2 = 10 \quad \text{and} \quad y_1 + y_2 = 0 \quad \text{(Equation 1)}
\]

From the midpoint of \( BC \), we have:

\[
\frac{x_2 + x_3}{2} = 5 \quad \text{and} \quad \frac{y_2 + y_3}{2} = 12
\]

This gives:

\[
x_2 + x_3 = 10 \quad \text{and} \quad y_2 + y_3 = 24 \quad \text{(Equation 2)}
\]

From the midpoint of \( CA \), we have:

\[
\frac{x_3 + x_1}{2} = 0 \quad \text{and} \quad \frac{y_3 + y_1}{2} = 12
\]

This gives:

\[
x_3 + x_1 = 0 \quad \text{and} \quad y_3 + y_1 = 24 \quad \text{(Equation 3)}
\]

### Step 2: Solve the system of equations
Now, we solve this system of equations.

1. From Equation 1: \( x_1 + x_2 = 10 \) and \( y_1 + y_2 = 0 \)
2. From Equation 2: \( x_2 + x_3 = 10 \) and \( y_2 + y_3 = 24 \)
3. From Equation 3: \( x_3 + x_1 = 0 \) and \( y_3 + y_1 = 24 \)

#### Solving for \( x_1, x_2, x_3 \):
- From Equation 3: \( x_3 = -x_1 \)
- Substitute \( x_3 = -x_1 \) into Equation 2: \( x_2 - x_1 = 10 \), which gives \( x_2 = x_1 + 10 \).
- Substitute \( x_2 = x_1 + 10 \) into Equation 1: \( x_1 + (x_1 + 10) = 10 \), which simplifies to \( 2x_1 + 10 = 10 \), so \( 2x_1 = 0 \), hence \( x_1 = 0 \).
- Therefore, \( x_2 = 10 \) and \( x_3 = 0 \).

#### Solving for \( y_1, y_2, y_3 \):
- From Equation 1: \( y_1 + y_2 = 0 \), so \( y_2 = -y_1 \).
- Substitute \( y_2 = -y_1 \) into Equation 2: \( -y_1 + y_3 = 24 \), which gives \( y_3 = y_1 + 24 \).
- Substitute \( y_3 = y_1 + 24 \) into Equation 3: \( y_1 + (y_1 + 24) = 24 \), which simplifies to \( 2y_1 + 24 = 24 \), so \( 2y_1 = 0 \), hence \( y_1 = 0 \).
- Therefore, \( y_2 = 0 \) and \( y_3 = 24 \).

### Step 3: Find the orthocenter
The vertices of the triangle are \( A(0, 0) \), \( B(10, 0) \), and \( C(0, 24) \).

The orthocenter of a triangle is the point where the altitudes (perpendiculars from the vertices to the opposite sides) intersect.

Since the triangle has a vertical side \( BC \) and a horizontal side \( AB \), the altitudes from \( A \) and \( C \) are along the x-axis and y-axis, respectively. These altitudes intersect at the origin.

Thus, the orthocenter is \( (0, 0) \).

### Final Answer:
The orthocenter of the triangle is \( (0, 0) \), so the correct answer is:

**A. (0, 0)**