To find the equation of the line BC in triangle ABC, where A is the point (-4, 5), E is (4, 1), and F is (-1, -4), we will follow these steps:
### Step 1: Find the slope of line AB
The coordinates of point A are (-4, 5) and we need to find the slope of line AB. We can assume point B has coordinates (x_B, y_B).
The slope of line AB is given by:
\[
\text{slope of AB} = \frac{y_B - 5}{x_B + 4}
\]
### Step 2: Find the slope of line AE
The coordinates of point E are (4, 1). The slope of line AE can be calculated as:
\[
\text{slope of AE} = \frac{1 - 5}{4 + 4} = \frac{-4}{8} = -\frac{1}{2}
\]
### Step 3: Find the slope of line OE
The slope of line OE is the negative reciprocal of the slope of line AE:
\[
\text{slope of OE} = 2
\]
### Step 4: Write the equation of line OE
Using point E (4, 1) and the slope of line OE (2), we can write the equation of line OE:
\[
y - 1 = 2(x - 4)
\]
Expanding this gives:
\[
y - 1 = 2x - 8 \implies y = 2x - 7
\]
### Step 5: Find the slope of line FO
Using point F (-1, -4) and the slope of line AB (which we will calculate later), we can find the slope of line FO. The slope of line AB was calculated earlier as:
\[
\text{slope of AB} = -3
\]
Thus, the slope of line FO is the negative reciprocal of this slope:
\[
\text{slope of FO} = \frac{1}{3}
\]
### Step 6: Write the equation of line FO
Using point F (-1, -4) and the slope of line FO (1/3), we can write:
\[
y + 4 = \frac{1}{3}(x + 1)
\]
Expanding this gives:
\[
y + 4 = \frac{1}{3}x + \frac{1}{3} \implies 3y + 12 = x + 1 \implies x - 3y - 11 = 0
\]
### Step 7: Find the intersection point O of lines OE and FO
Now we have two equations:
1. \(y = 2x - 7\) (Equation of OE)
2. \(x - 3y - 11 = 0\) (Equation of FO)
Substituting the first equation into the second:
\[
x - 3(2x - 7) - 11 = 0
\]
Solving this gives:
\[
x - 6x + 21 - 11 = 0 \implies -5x + 10 = 0 \implies x = 2
\]
Substituting \(x = 2\) back into \(y = 2x - 7\):
\[
y = 2(2) - 7 = 4 - 7 = -3
\]
Thus, the coordinates of point O are (2, -3).
### Step 8: Find the slope of line AD
The slope of line AD can be calculated as:
\[
\text{slope of AD} = \frac{5 - (-3)}{-4 - 2} = \frac{8}{-6} = -\frac{4}{3}
\]
### Step 9: Find the slope of line BC
The slope of line BC is the negative reciprocal of the slope of line AD:
\[
\text{slope of BC} = \frac{3}{4}
\]
### Step 10: Write the equation of line BC
Using point C (which we will find using the coordinates of A and E) and the slope of line BC, we can write:
\[
y - y_C = \frac{3}{4}(x - x_C)
\]
### Step 11: Find the coordinates of point C
To find point C, we can use the coordinates of A and E to find the equation of line AC:
Using points A (-4, 5) and E (4, 1):
\[
\text{slope of AC} = \frac{1 - 5}{4 + 4} = \frac{-4}{8} = -\frac{1}{2}
\]
The equation of line AC is:
\[
y - 5 = -\frac{1}{2}(x + 4)
\]
Expanding gives:
\[
y - 5 = -\frac{1}{2}x - 2 \implies y = -\frac{1}{2}x + 3
\]
### Step 12: Solve for the intersection of lines AC and FO
Now we can solve:
1. \(y = -\frac{1}{2}x + 3\)
2. \(y = \frac{1}{3}x - 4\)
Setting them equal:
\[
-\frac{1}{2}x + 3 = \frac{1}{3}x - 4
\]
Clearing fractions by multiplying through by 6:
\[
-3x + 18 = 2x - 24 \implies 5x = 42 \implies x = \frac{42}{5}
\]
Substituting \(x\) back to find \(y\):
\[
y = -\frac{1}{2}(\frac{42}{5}) + 3 = -\frac{21}{5} + \frac{15}{5} = -\frac{6}{5}
\]
### Step 13: Write the final equation of line BC
Using point C \((\frac{42}{5}, -\frac{6}{5})\) and the slope \(\frac{3}{4}\):
\[
y + \frac{6}{5} = \frac{3}{4}(x - \frac{42}{5})
\]
Expanding and simplifying gives:
\[
4y + \frac{24}{5} = 3x - \frac{126}{5} \implies 3x - 4y - \frac{150}{5} = 0 \implies 3x - 4y - 30 = 0
\]
Thus, the final equation of line BC is:
\[
3x - 4y - 30 = 0
\]
