If `I_(1), I_(2), I_(3)` are excentres of the triangle with vertices (0, 0), (5, 12), (16, 12) then the orthocentre of `DeltaI_(1)I_(2)I_(3)` is

Answer

To find the orthocenter of the triangle formed by the excenters \( I_1, I_2, I_3 \) of triangle \( ABC \) with vertices \( A(0, 0) \), \( B(5, 12) \), and \( C(16, 12) \), we can use the property that the orthocenter of triangle \( I_1 I_2 I_3 \) is equal to the incenter of triangle \( ABC \).

### Step-by-step Solution:

1. **Calculate the lengths of the sides of triangle \( ABC \)**:
- \( a = BC \)
- \( b = AC \)
- \( c = AB \)

Using the distance formula:
- \( a = BC = \sqrt{(16 - 5)^2 + (12 - 12)^2} = \sqrt{11^2} = 11 \)
- \( b = AC = \sqrt{(16 - 0)^2 + (12 - 0)^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \)
- \( c = AB = \sqrt{(5 - 0)^2 + (12 - 0)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)

2. **Use the formula for the incenter coordinates**:
The coordinates of the incenter \( I \) are given by:
\[ I_x = \frac{a \cdot x_A + b \cdot x_B + c \cdot x_C}{a + b + c} \]
\[ I_y = \frac{a \cdot y_A + b \cdot y_B + c \cdot y_C}{a + b + c} \]

Substituting the values:
- \( x_A = 0, y_A = 0 \)
- \( x_B = 5, y_B = 12 \)
- \( x_C = 16, y_C = 12 \)

Calculate \( I_x \):
\[ I_x = \frac{11 \cdot 0 + 20 \cdot 5 + 13 \cdot 16}{11 + 20 + 13} = \frac{0 + 100 + 208}{44} = \frac{308}{44} = 7 \]

Calculate \( I_y \):
\[ I_y = \frac{11 \cdot 0 + 20 \cdot 12 + 13 \cdot 12}{11 + 20 + 13} = \frac{0 + 240 + 156}{44} = \frac{396}{44} = 9 \]

3. **Conclusion**:
The coordinates of the incenter \( I \) are \( (7, 9) \). Therefore, the orthocenter of triangle \( I_1 I_2 I_3 \) is also \( (7, 9) \).

### Final Answer:
The orthocenter of triangle \( I_1 I_2 I_3 \) is \( (7, 9) \).