Let A,B,C be angles of triangles with vertex `A -= (4,-1)` and internal angular bisectors of angles B and C be `x - 1 = 0` and `x - y - 1 = 0` respectively.
Slope of BC is

Options

A. `1//2`
B. 2
C. 3
D. 12

Explanation

To find the slope of line segment BC in triangle ABC, where point A is given as \( A(4, -1) \) and the internal angular bisectors of angles B and C are given by the equations \( x - 1 = 0 \) and \( x - y - 1 = 0 \), we can follow these steps:

### Step 1: Identify the equations of the angular bisectors
The angular bisector of angle B is given by the equation \( x - 1 = 0 \). This is a vertical line that passes through the point \( (1, y) \) for any value of \( y \).

The angular bisector of angle C is given by the equation \( x - y - 1 = 0 \). We can rearrange this to find the slope:
\[ y = x - 1 \]
This line has a slope of 1.

### Step 2: Find the intersection point of the bisectors
To find the coordinates of point B (where the bisector of angle B intersects the bisector of angle C), we substitute \( x = 1 \) into the equation of the bisector of angle C:
\[ y = 1 - 1 = 0 \]
Thus, point B is at \( B(1, 0) \).

### Step 3: Find the image of point A with respect to the bisector of angle C
To find the image of point A with respect to the line \( x - y - 1 = 0 \), we can use the formula for finding the image of a point across a line. The line can be rewritten in the standard form \( Ax + By + C = 0 \) where \( A = 1, B = -1, C = -1 \).

Using the formula for the image of point \( (x_1, y_1) \):
\[ \text{Image} = \left( \frac{x_1(B^2 - A^2) - 2AB y_1 - 2AC}{A^2 + B^2}, \frac{y_1(A^2 - B^2) - 2AB x_1 - 2BC}{A^2 + B^2} \right) \]
Substituting \( A = 1, B = -1, C = -1 \) and \( (x_1, y_1) = (4, -1) \):
\[ \text{Image} = \left( \frac{4((-1)^2 - 1^2) - 2(1)(-1)(-1) - 2(1)(-1)}{1^2 + (-1)^2}, \frac{-1(1^2 - (-1)^2) - 2(1)(-1)(4) - 2(-1)(-1)}{1^2 + (-1)^2} \right) \]
Calculating this gives us the image point.

### Step 4: Find the coordinates of point C
Similarly, we can find the image of point A with respect to the line \( x - 1 = 0 \) using the same method.

### Step 5: Calculate the slope of line segment BC
Once we have the coordinates of points B and C, we can find the slope of line segment BC using the formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
where \( (x_1, y_1) \) are the coordinates of point B and \( (x_2, y_2) \) are the coordinates of point C.

### Final Answer
After performing the calculations, we find that the slope of line segment BC is \( 2 \).

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