In the triangle ABC, vertices A, B, C are `(1, 2), (3, 1), (-1, 6)` respectively. If the internal angle bisector of `angleBAC` meets BC at D, then the coordinates of D are `((5)/(3),(8)/(3))`

Options

A. `(1, (7)/(2))`
B. `((1)/(3),(13)/(3))`
C. `(2, (3)/(4))`
D. `(2, (3)/(4))`

Answer

To find the coordinates of point D, where the internal angle bisector of angle BAC meets side BC in triangle ABC with vertices A(1, 2), B(3, 1), and C(-1, 6), we can use the Angle Bisector Theorem and the Section Formula.

### Step-by-Step Solution:

1. **Identify the Vertices**:
- A(1, 2)
- B(3, 1)
- C(-1, 6)

2. **Calculate the Lengths of Sides AB and AC**:
- Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 1)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]
- For AC:
\[ AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-1 - 1)^2 + (6 - 2)^2} = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

3. **Find the Ratio of AB to AC**:
- The ratio \( \frac{AB}{AC} = \frac{\sqrt{5}}{2\sqrt{5}} = \frac{1}{2} \).
- By the Angle Bisector Theorem, this means \( \frac{BD}{DC} = \frac{1}{2} \).

4. **Apply the Section Formula**:
- Let D divide BC in the ratio 1:2. The coordinates of D can be found using the section formula:
\[ D\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \]
- Here, \( m = 1 \), \( n = 2 \), \( B(3, 1) \), and \( C(-1, 6) \).

5. **Calculate the Coordinates of D**:
- For the x-coordinate:
\[ x_D = \frac{1 \cdot (-1) + 2 \cdot 3}{1 + 2} = \frac{-1 + 6}{3} = \frac{5}{3} \]
- For the y-coordinate:
\[ y_D = \frac{1 \cdot 6 + 2 \cdot 1}{1 + 2} = \frac{6 + 2}{3} = \frac{8}{3} \]

6. **Final Coordinates of D**:
- Therefore, the coordinates of D are \( D\left(\frac{5}{3}, \frac{8}{3}\right) \).