If AD, BE and CF are the altitudes of `Delta ABC` whose vertex A is (-4,5). The coordinates of points E and F are (4,1) and (-1,-4), respectively. Equation of BC is

Options

A. 3x-4y+28=0
B. 4x+3y+28=0
C. 3x-4y-28=0
D. x+2y+7=0

Explanation

To find the equation of line BC in triangle ABC, we will follow these steps:

### Step 1: Identify the given points
We have the following points:
- Vertex A = (-4, 5)
- Point E (foot of altitude from B) = (4, 1)
- Point F (foot of altitude from C) = (-1, -4)

### Step 2: Find the coordinates of point B
To find the coordinates of point B, we need to find the intersection of lines AB and BE.

#### Equation of line AB:
1. **Find the slope of line AF**:
- Coordinates of A = (-4, 5) and F = (-1, -4)
- Slope (m) = (y2 - y1) / (x2 - x1) = (-4 - 5) / (-1 + 4) = -9 / 3 = -3

2. **Use point-slope form to find the equation of line AB**:
- Using point A (-4, 5) and slope -3:
- Equation: y - 5 = -3(x + 4)
- Rearranging gives: 3x + y + 7 = 0 (Equation 1)

#### Equation of line BE:
1. **Find the slope of line AE**:
- Coordinates of A = (-4, 5) and E = (4, 1)
- Slope (m) = (1 - 5) / (4 + 4) = -4 / 8 = -1/2

2. **Use point-slope form to find the equation of line BE**:
- Using point E (4, 1) and slope 2 (perpendicular to AE):
- Equation: y - 1 = 2(x - 4)
- Rearranging gives: 2x - y - 7 = 0 (Equation 2)

### Step 3: Solve the equations of AB and BE to find point B
Now we will solve Equation 1 and Equation 2 simultaneously:
1. From Equation 1: 3x + y + 7 = 0
2. From Equation 2: 2x - y - 7 = 0

Adding both equations:
- (3x + y + 7) + (2x - y - 7) = 0
- 5x = 0
- x = 0

Substituting x = 0 into Equation 1:
- 3(0) + y + 7 = 0
- y = -7

Thus, coordinates of point B are (0, -7).

### Step 4: Find the coordinates of point C
To find the coordinates of point C, we need to find the intersection of lines AC and CF.

#### Equation of line AC:
1. **Find the slope of line AE**:
- Slope (m) = (1 - 5) / (4 + 4) = -4 / 8 = -1/2

2. **Use point-slope form to find the equation of line AC**:
- Using point A (-4, 5) and slope -1/2:
- Equation: y - 5 = -1/2(x + 4)
- Rearranging gives: x + 2y - 6 = 0 (Equation 3)

#### Equation of line CF:
1. **Find the slope of line AF**:
- Slope (m) = (y2 - y1) / (x2 - x1) = (-4 - 5) / (-1 + 4) = -9 / 3 = -3

2. **Use point-slope form to find the equation of line CF**:
- Using point F (-1, -4) and slope 1/3 (perpendicular to AF):
- Equation: y + 4 = 1/3(x + 1)
- Rearranging gives: x - 3y + 11 = 0 (Equation 4)

### Step 5: Solve the equations of AC and CF to find point C
Now we will solve Equation 3 and Equation 4 simultaneously:
1. From Equation 3: x + 2y - 6 = 0
2. From Equation 4: x - 3y + 11 = 0

Subtracting Equation 3 from Equation 4:
- (x - 3y + 11) - (x + 2y - 6) = 0
- -5y + 17 = 0
- y = 17/5 = 3.4

Substituting y = 3.4 into Equation 3:
- x + 2(3.4) - 6 = 0
- x + 6.8 - 6 = 0
- x = -0.8

Thus, coordinates of point C are (-0.8, 3.4).

### Step 6: Find the equation of line BC
1. **Find the slope of line BC**:
- Coordinates of B = (0, -7) and C = (-0.8, 3.4)
- Slope (m) = (3.4 + 7) / (-0.8 - 0) = 10.4 / -0.8 = -13

2. **Use point-slope form to find the equation of line BC**:
- Using point B (0, -7):
- Equation: y + 7 = -13(x - 0)
- Rearranging gives: 13x + y + 7 = 0

Thus, the equation of line BC is:
**13x + y + 7 = 0**