In an equilateral triangle having vertex A(8,9), B(-4, 3), C(h, k) sum of all possible h and k

Options

A. 13
B. 16
C. 26
D. 43

Explanation

To solve the problem of finding the sum of all possible values of \( h \) and \( k \) for the vertex \( C(h, k) \) of an equilateral triangle with vertices \( A(8, 9) \) and \( B(-4, 3) \), we can follow these steps:

### Step 1: Find the Midpoint of AB
The midpoint \( O \) of segment \( AB \) can be calculated using the midpoint formula:
\[ O = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
where \( A(8, 9) \) and \( B(-4, 3) \).

Calculating:
\[ O_x = \frac{8 + (-4)}{2} = \frac{4}{2} = 2 \]
\[ O_y = \frac{9 + 3}{2} = \frac{12}{2} = 6 \]
Thus, the coordinates of point \( O \) are \( (2, 6) \).

**Hint**: Use the midpoint formula to find the center of the line segment connecting points A and B.

### Step 2: Set Up Equations for Coordinates of C
Since \( O \) is also the midpoint of \( C(h, k) \) and the other vertex (let's denote it as \( D \)), we can set up the following equations:
\[ \frac{h_1 + h_2}{2} = 2 \quad \text{(1)} \]
\[ \frac{k_1 + k_2}{2} = 6 \quad \text{(2)} \]
where \( (h_1, k_1) \) and \( (h_2, k_2) \) are the coordinates of the two possible positions of point \( C \).

**Hint**: Remember that the midpoint of a segment is the average of the coordinates of its endpoints.

### Step 3: Solve for h and k
From equations (1) and (2), we can express \( h_1 + h_2 \) and \( k_1 + k_2 \):
\[ h_1 + h_2 = 4 \quad \text{(3)} \]
\[ k_1 + k_2 = 12 \quad \text{(4)} \]

### Step 4: Find the Total Sum
Now, we need to find the sum of all possible values of \( h \) and \( k \):
\[ h_1 + h_2 + k_1 + k_2 = 4 + 12 = 16 \]

### Conclusion
Thus, the sum of all possible values of \( h \) and \( k \) is \( 16 \).

**Final Answer**: The sum of all possible values of \( h \) and \( k \) is \( 16 \).

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