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

Options

A. (0,0)
B. (10,0)
C. (0,24)
D. `(13/3,8)`

Answer

To find the orthocenter of the triangle given the midpoints of its sides, we can follow these steps:

### Step 1: Identify the vertices of the triangle
The midpoints of the triangle are given as:
- M1 = (5, 0)
- M2 = (5, 12)
- M3 = (0, 12)

Let’s denote the vertices of the triangle as A, B, and C. The midpoints of the sides of triangle ABC can be expressed in terms of the coordinates of the vertices:
- M1 = (A + B)/2
- M2 = (B + C)/2
- M3 = (C + A)/2

### Step 2: Set up equations for the vertices
From the midpoint coordinates, we can set up the following equations:

1. For M1 = (5, 0):
\[ \frac{x_A + x_B}{2} = 5 \quad \text{and} \quad \frac{y_A + y_B}{2} = 0 \]
This gives us:
\[ x_A + x_B = 10 \quad (1) \]
\[ y_A + y_B = 0 \quad (2) \]

2. For M2 = (5, 12):
\[ \frac{x_B + x_C}{2} = 5 \quad \text{and} \quad \frac{y_B + y_C}{2} = 12 \]
This gives us:
\[ x_B + x_C = 10 \quad (3) \]
\[ y_B + y_C = 24 \quad (4) \]

3. For M3 = (0, 12):
\[ \frac{x_C + x_A}{2} = 0 \quad \text{and} \quad \frac{y_C + y_A}{2} = 12 \]
This gives us:
\[ x_C + x_A = 0 \quad (5) \]
\[ y_C + y_A = 24 \quad (6) \]

### Step 3: Solve the equations
From equation (5), we can express \(x_A\) in terms of \(x_C\):
\[ x_A = -x_C \quad (7) \]

Substituting equation (7) into equation (1):
\[ -x_C + x_B = 10 \quad (8) \]

Now, substituting equation (7) into equation (6):
\[ y_C - y_A = 24 \quad (9) \]

Now we have three equations (3), (8), and (9) to solve for \(x_B\), \(x_C\), \(y_A\), \(y_B\), and \(y_C\).

From equation (3):
\[ x_B + x_C = 10 \quad (3) \]

Substituting \(x_B\) from equation (8) into (3):
\[ 10 + x_C = 10 \implies x_C = 0 \quad (10) \]

Substituting \(x_C = 0\) back into equation (7):
\[ x_A = -0 = 0 \quad (11) \]

Substituting \(x_C = 0\) into equation (3):
\[ x_B = 10 \quad (12) \]

Now we have \(x_A = 0\), \(x_B = 10\), and \(x_C = 0\).

### Step 4: Solve for y-coordinates
Using equations (2) and (4):
From (2):
\[ y_A + y_B = 0 \quad (2) \]

From (4):
\[ y_B + y_C = 24 \quad (4) \]

Substituting \(y_B = -y_A\) from (2) into (4):
\[ -y_A + y_C = 24 \quad (13) \]

Now substituting \(y_C = 24 + y_A\) into (6):
\[ (24 + y_A) + y_A = 24 \]
\[ 2y_A + 24 = 24 \implies 2y_A = 0 \implies y_A = 0 \quad (14) \]

From (2):
\[ y_B = 0 \quad (15) \]

Substituting \(y_A = 0\) into (4):
\[ 0 + y_C = 24 \implies y_C = 24 \quad (16) \]

### Step 5: Determine the coordinates of the vertices
Now we have the coordinates of the vertices:
- A = (0, 0)
- B = (10, 0)
- C = (0, 24)

### Step 6: Find the orthocenter
In a right triangle, the orthocenter is located at the vertex where the right angle is formed. Since triangle ABC has a right angle at vertex A (0, 0), the orthocenter is at point A.

Thus, the orthocenter of the triangle is:
\[ \text{Orthocenter} = (0, 0) \]

### Final Answer:
The orthocenter of the triangle is (0, 0).

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