`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.` find the equation of 'BC'

Options

A. 7x+3y-4=0
B. 7x+3y+4=0
C. 7x-3y+4=0
D. 7x-3y-4=0

Explanation

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

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

### Step 2: Determine the equation of the angle bisector
The equation of the angle bisector of angle ABC is given as:
\[ x + y = 2 \]

### Step 3: Find the image of point A with respect to the angle bisector
Let D be the image of point A with respect to line BC. We can use the formula for finding the image of a point with respect to a line.

The general form of the line is \( Ax + By + C = 0 \). For the line \( x + y - 2 = 0 \):
- A = 1, B = 1, C = -2

Using the formula for the coordinates of the image \( D(h, k) \):
\[ \frac{h - x_1}{A} = \frac{k - y_1}{B} = \frac{-2(Ax_1 + By_1 + C)}{A^2 + B^2} \]
Substituting \( A(1, 3) \) into the formula:
\[ \frac{h - 1}{1} = \frac{k - 3}{1} = \frac{-2(1 \cdot 1 + 1 \cdot 3 - 2)}{1^2 + 1^2} \]
Calculating the right side:
\[ = \frac{-2(1 + 3 - 2)}{2} = \frac{-2 \cdot 2}{2} = -2 \]
So we have:
\[ h - 1 = -2 \quad \text{and} \quad k - 3 = -2 \]
Solving these gives:
\[ h = -1 \quad \text{and} \quad k = 1 \]
Thus, the coordinates of point D are:
\[ D(-1, 1) \]

### Step 4: Use points D and C to find the equation of line BC
Now we will find the equation of line BC using the coordinates of points D and C:
- D(-1, 1)
- C(-2/5, -2/5)

Using the two-point form of the equation of a line:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting \( (x_1, y_1) = (-1, 1) \) and \( (x_2, y_2) = \left(-\frac{2}{5}, -\frac{2}{5}\right) \):
\[ y - 1 = \frac{-\frac{2}{5} - 1}{-\frac{2}{5} + 1}(x + 1) \]
Calculating the slope:
\[ = \frac{-\frac{2}{5} - \frac{5}{5}}{-\frac{2}{5} + \frac{5}{5}} = \frac{-\frac{7}{5}}{\frac{3}{5}} = -\frac{7}{3} \]
Thus, the equation becomes:
\[ y - 1 = -\frac{7}{3}(x + 1) \]
Expanding this:
\[ y - 1 = -\frac{7}{3}x - \frac{7}{3} \]
Adding 1 to both sides:
\[ y = -\frac{7}{3}x - \frac{7}{3} + 1 \]
Converting 1 to a fraction:
\[ y = -\frac{7}{3}x - \frac{7}{3} + \frac{3}{3} = -\frac{7}{3}x - \frac{4}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 3y = -7x - 4 \]
Rearranging gives:
\[ 7x + 3y + 4 = 0 \]

### Final Answer
The equation of line BC is:
\[ 7x + 3y + 4 = 0 \]