The coordinates of the orthocenter of the triangle that has the coordinates of midpoint of its sides as ( 0,0) , ( 1,2) and ( - 6,3) is

Options

A. ( 0,0)
B. ( - 4,5)
C. ( -5,5)
D. ( -4,4)

Answer

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

### Step 1: Identify the Midpoints
The midpoints of the triangle's sides are given as:
- \( M_1 = (0, 0) \)
- \( M_2 = (1, 2) \)
- \( M_3 = (-6, 3) \)

### Step 2: Use the Midpoint Formula
The coordinates of the vertices of the triangle can be derived from the midpoints. If \( A, B, C \) are the vertices of the triangle, then the midpoints can be expressed as:
- \( M_1 = \frac{A + B}{2} \)
- \( M_2 = \frac{B + C}{2} \)
- \( M_3 = \frac{C + A}{2} \)

### Step 3: Set Up the Equations
From the midpoint coordinates, we can set up the following equations:
1. \( 0 = \frac{A_x + B_x}{2} \) (from \( M_1 \))
2. \( 1 = \frac{B_x + C_x}{2} \) (from \( M_2 \))
3. \( -6 = \frac{C_x + A_x}{2} \) (from \( M_3 \))

And similarly for the y-coordinates:
1. \( 0 = \frac{A_y + B_y}{2} \)
2. \( 2 = \frac{B_y + C_y}{2} \)
3. \( 3 = \frac{C_y + A_y}{2} \)

### Step 4: Solve for the Coordinates of Vertices
From the x-coordinates:
1. \( A_x + B_x = 0 \) (Equation 1)
2. \( B_x + C_x = 2 \) (Equation 2)
3. \( C_x + A_x = -12 \) (Equation 3)

From Equation 1, we can express \( B_x \) in terms of \( A_x \):
\[ B_x = -A_x \]

Substituting \( B_x \) into Equation 2:
\[ -A_x + C_x = 2 \]
Thus, \( C_x = A_x + 2 \).

Now substituting \( B_x \) and \( C_x \) into Equation 3:
\[ (A_x + 2) + A_x = -12 \]
\[ 2A_x + 2 = -12 \]
\[ 2A_x = -14 \]
\[ A_x = -7 \]

Now substituting back to find \( B_x \) and \( C_x \):
\[ B_x = -(-7) = 7 \]
\[ C_x = -7 + 2 = -5 \]

Now we have:
- \( A_x = -7 \)
- \( B_x = 7 \)
- \( C_x = -5 \)

For the y-coordinates:
1. \( A_y + B_y = 0 \) (Equation 4)
2. \( B_y + C_y = 4 \) (Equation 5)
3. \( C_y + A_y = 6 \) (Equation 6)

From Equation 4, we can express \( B_y \) in terms of \( A_y \):
\[ B_y = -A_y \]

Substituting \( B_y \) into Equation 5:
\[ -A_y + C_y = 4 \]
Thus, \( C_y = A_y + 4 \).

Now substituting \( B_y \) and \( C_y \) into Equation 6:
\[ (A_y + 4) + A_y = 6 \]
\[ 2A_y + 4 = 6 \]
\[ 2A_y = 2 \]
\[ A_y = 1 \]

Now substituting back to find \( B_y \) and \( C_y \):
\[ B_y = -1 \]
\[ C_y = 1 + 4 = 5 \]

Now we have:
- \( A = (-7, 1) \)
- \( B = (7, -1) \)
- \( C = (-5, 5) \)

### Step 5: Find the Orthocenter
The orthocenter of a triangle formed by vertices \( A, B, C \) can be found using the property that in a right triangle, the orthocenter is at the vertex where the right angle is formed.

To find the slopes of the sides:
- Slope of \( AB \): \( \frac{-1 - 1}{7 - (-7)} = \frac{-2}{14} = -\frac{1}{7} \)
- Slope of \( BC \): \( \frac{5 - (-1)}{-5 - 7} = \frac{6}{-12} = -\frac{1}{2} \)
- Slope of \( CA \): \( \frac{1 - 5}{-7 - (-5)} = \frac{-4}{-2} = 2 \)

Since the slopes of \( AB \) and \( BC \) are negative reciprocals, \( \angle ABC \) is a right angle. Therefore, the orthocenter is at point \( C \).

### Final Answer
The coordinates of the orthocenter of the triangle are:
\[ \text{Orthocenter} = (-5, 5) \]