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 follow these steps:

### Step 1: Identify the coordinates of points A, B, and C.
- A = (1, 2)
- B = (3, 1)
- C = (-1, 6)

### Step 2: Calculate the lengths of sides BC and AC.
- To find the length of side BC, we use the distance formula:
\[ BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \sqrt{((-1) - 3)^2 + (6 - 1)^2} \]
\[ = \sqrt{(-4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41} \]

- To find the length of side AC:
\[ AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} = \sqrt{((-1) - 1)^2 + (6 - 2)^2} \]
\[ = \sqrt{(-2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

### Step 3: Use the angle bisector theorem.
According to the angle bisector theorem, the ratio in which D divides BC is given by:
\[ \frac{BD}{DC} = \frac{AC}{AB} \]

- First, we need to calculate the length of side AB:
\[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(3 - 1)^2 + (1 - 2)^2} \]
\[ = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

- Now we can find the ratio:
\[ \frac{BD}{DC} = \frac{AC}{AB} = \frac{2\sqrt{5}}{\sqrt{5}} = 2:1 \]

### Step 4: Find the coordinates of point D using the section formula.
Let D divide BC in the ratio 2:1. Using the section formula:
\[ D\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \]
where \(m = 2\), \(n = 1\), \(B(3, 1)\), and \(C(-1, 6)\):
\[ D_x = \frac{2 \cdot (-1) + 1 \cdot 3}{2 + 1} = \frac{-2 + 3}{3} = \frac{1}{3} \]
\[ D_y = \frac{2 \cdot 6 + 1 \cdot 1}{2 + 1} = \frac{12 + 1}{3} = \frac{13}{3} \]

### Step 5: Final coordinates of D.
Thus, the coordinates of point D are:
\[ D\left(\frac{5}{3}, \frac{8}{3}\right) \]