`A(1,3)and c(-2/5,-2/5)`are the vertices of a `DeltaABCand`the equation of the angle bisector of `/_ABC` is `x+y=2.`

Options

A. 3x+7y=24
B. 3x+7y+24=0
C. 13x+7y+8=0
D. 13x-7y+8=0

Explanation

To find the equation of line BC given the vertices A(1, 3) and C(-2/5, -2/5), and the angle bisector of angle ABC is given by the equation x + y = 2, we can follow these steps:

### Step 1: Identify the coordinates of points A and C
We have:
- Point A = (1, 3)
- Point C = (-2/5, -2/5)

### Step 2: Find the slope of the angle bisector
The equation of the angle bisector is given as x + y = 2. We can rewrite this in slope-intercept form (y = mx + b):
- y = -x + 2
Thus, the slope (m) of the angle bisector is -1.

### Step 3: Find the coordinates of point B
Let’s denote the coordinates of point B as (h, k). Since the angle bisector divides the angle at B into two equal angles, we can use the section formula to find the coordinates of point D on line AC such that AD/DB = AC/BC.

### Step 4: Find the coordinates of point D
Using the section formula, we can find the coordinates of point D. The coordinates of D can be expressed as:
\[ D\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \]
where (x1, y1) = A(1, 3) and (x2, y2) = C(-2/5, -2/5). Let’s assume m:n is the ratio in which D divides AC.

### Step 5: Set up the equations for point D
Using the coordinates of A and C:
\[ h = \frac{m \cdot (-\frac{2}{5}) + n \cdot 1}{m+n} \]
\[ k = \frac{m \cdot (-\frac{2}{5}) + n \cdot 3}{m+n} \]
Since we need to find the coordinates of point D that lies on the angle bisector, we can substitute k = -h + 2 into these equations.

### Step 6: Solve for h and k
From the angle bisector equation:
\[ k = -h + 2 \]
Substituting this into the equations for h and k, we can solve for h and k.

### Step 7: Find the equation of line BC
Once we have the coordinates of point B (h, k), we can use the two-point form of the line equation to find the equation of line BC:
\[ y - k = \frac{y_2 - y_1}{x_2 - x_1}(x - h) \]
where (x1, y1) = B(h, k) and (x2, y2) = C(-\frac{2}{5}, -\frac{2}{5}).

### Step 8: Simplify the equation
After substituting the values and simplifying, we will arrive at the final equation of line BC.

### Final Equation
The final equation of line BC will be in the form:
\[ Ax + By + C = 0 \]
where A, B, and C are constants derived from the previous steps.