The vertices of a triangle are (1, 2), (h, -3) and (-4, k). Find the value of `sqrt({(h+k)^(2)+(h+3k)^(2)})`. If the centroid of the triangle be at point (5, -1).

Answer

To solve the problem, we need to find the value of \( \sqrt{(h+k)^2 + (h+3k)^2} \) given that the centroid of the triangle formed by the vertices \( (1, 2) \), \( (h, -3) \), and \( (-4, k) \) is at the point \( (5, -1) \).

### Step-by-step Solution:

1. **Understanding the Centroid Formula**:
The centroid \( G \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

2. **Substituting the Given Points**:
For our triangle, the vertices are \( (1, 2) \), \( (h, -3) \), and \( (-4, k) \). Thus, we can write:
\[ G = \left( \frac{1 + h - 4}{3}, \frac{2 - 3 + k}{3} \right) \]
This simplifies to:
\[ G = \left( \frac{h - 3}{3}, \frac{k - 1}{3} \right) \]

3. **Setting Up the Equations**:
We know from the problem that the centroid \( G \) is \( (5, -1) \). Therefore, we can set up the following equations:
\[ \frac{h - 3}{3} = 5 \quad \text{(1)} \]
\[ \frac{k - 1}{3} = -1 \quad \text{(2)} \]

4. **Solving for \( h \)**:
From equation (1):
\[ h - 3 = 15 \implies h = 18 \]

5. **Solving for \( k \)**:
From equation (2):
\[ k - 1 = -3 \implies k = -2 \]

6. **Calculating \( h + k \) and \( h + 3k \)**:
Now we can find \( h + k \) and \( h + 3k \):
\[ h + k = 18 - 2 = 16 \]
\[ h + 3k = 18 + 3(-2) = 18 - 6 = 12 \]

7. **Finding the Value of the Expression**:
We need to compute:
\[ \sqrt{(h+k)^2 + (h+3k)^2} = \sqrt{16^2 + 12^2} \]
Calculating the squares:
\[ 16^2 = 256, \quad 12^2 = 144 \]
Adding these:
\[ 256 + 144 = 400 \]
Finally, taking the square root:
\[ \sqrt{400} = 20 \]

### Final Answer:
The value of \( \sqrt{(h+k)^2 + (h+3k)^2} \) is \( 20 \).

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