The co-ordinates of the vertices of `DeltaABC` are `A(-2,4)`, `B(5,5)` and `C(4,-2)`. The equation of the bisector of `/_A` is :

Options

A. `x+3y=10`
B. `x-3y=10`
C. `3x+y=10`
D. `3x-y=10`

Answer

To find the equation of the angle bisector of angle A in triangle ABC with vertices A(-2, 4), B(5, 5), and C(4, -2), we can follow these steps:

### Step 1: Find the lengths of sides AB and AC
Using the distance formula, we can calculate the lengths of sides AB and AC.

- **Distance AB**:
\[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(5 - (-2))^2 + (5 - 4)^2} = \sqrt{(5 + 2)^2 + (1)^2} = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \]

- **Distance AC**:
\[ AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} = \sqrt{(4 - (-2))^2 + (-2 - 4)^2} = \sqrt{(4 + 2)^2 + (-6)^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]

### Step 2: Use the Angle Bisector Theorem
According to the Angle Bisector Theorem, the ratio of the segments created by the angle bisector is equal to the ratio of the lengths of the opposite sides:
\[ \frac{BP}{PC} = \frac{AB}{AC} = \frac{5\sqrt{2}}{6\sqrt{2}} = \frac{5}{6} \]

### Step 3: Find the coordinates of point P using the section formula
Point P divides segment BC in the ratio 5:6. Using the section formula:
\[ P\left(\frac{m \cdot x_2 + n \cdot x_1}{m+n}, \frac{m \cdot y_2 + n \cdot y_1}{m+n}\right) \]
where \(m = 5\) and \(n = 6\), and coordinates of B(5, 5) and C(4, -2):
\[ P_x = \frac{5 \cdot 4 + 6 \cdot 5}{5 + 6} = \frac{20 + 30}{11} = \frac{50}{11} \]
\[ P_y = \frac{5 \cdot (-2) + 6 \cdot 5}{5 + 6} = \frac{-10 + 30}{11} = \frac{20}{11} \]
Thus, the coordinates of point P are \(\left(\frac{50}{11}, \frac{20}{11}\right)\).

### Step 4: Find the slope of line AP
The slope \(m\) of line AP can be calculated as follows:
\[ m = \frac{y_P - y_A}{x_P - x_A} = \frac{\frac{20}{11} - 4}{\frac{50}{11} - (-2)} = \frac{\frac{20}{11} - \frac{44}{11}}{\frac{50}{11} + \frac{22}{11}} = \frac{\frac{-24}{11}}{\frac{72}{11}} = -\frac{24}{72} = -\frac{1}{3} \]

### Step 5: Write the equation of line AP
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = -\frac{1}{3}\), \(x_1 = -2\), and \(y_1 = 4\):
\[ y - 4 = -\frac{1}{3}(x + 2) \]
Multiplying through by 3 to eliminate the fraction:
\[ 3(y - 4) = -(x + 2) \]
\[ 3y - 12 = -x - 2 \]
Rearranging gives:
\[ x + 3y = 10 \]

### Final Answer
The equation of the angle bisector of angle A is:
\[ x + 3y = 10 \]