BE and CF are two medians of `DeltaABC` whose vertex A is (1,3). The equation of BE is `x-2y+1=0` and CF is y-1=0.
Determine the following:
The co-ordinates of points B,C and centroid G. The equations of lines AB and AC.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the coordinates of points B and C using the equations of the medians.

Given:
- Vertex A = (1, 3)
- Median BE: \( x - 2y + 1 = 0 \)
- Median CF: \( y - 1 = 0 \)

From the equation of median CF, we can find the y-coordinate of point C directly:
\[ y - 1 = 0 \Rightarrow y = 1 \]
Thus, point C has coordinates \( (k, 1) \) where \( k \) is unknown.

### Step 2: Substitute y = 1 into the equation of median BE to find the x-coordinate of point B.

Substituting \( y = 1 \) into the equation of BE:
\[ x - 2(1) + 1 = 0 \]
\[ x - 2 + 1 = 0 \]
\[ x - 1 = 0 \Rightarrow x = 1 \]
Thus, point B has coordinates \( (1, h) \) where \( h \) is unknown.

### Step 3: Use the centroid formula to find the coordinates of points B and C.

The centroid G of triangle ABC is given by:
\[ G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right) \]

Since G is the intersection of the medians, we can find its coordinates by solving the equations of the medians.

From the video transcript, we have:
- Centroid G = (1, 1)

Using the centroid formula:
\[ 1 = \frac{1 + 1 + k}{3} \]
\[ 3 = 2 + k \Rightarrow k = 1 \]

Now we can find the y-coordinate of point B using the same formula:
\[ 1 = \frac{3 + h + 1}{3} \]
\[ 3 = 4 + h \Rightarrow h = -1 \]

Thus, we have:
- Point B = (1, -1)
- Point C = (1, 1)

### Step 4: Find the coordinates of the centroid G.

The coordinates of the centroid G are already found to be \( (1, 1) \).

### Step 5: Determine the equations of lines AB and AC.

**Equation of line AB:**
Using points A (1, 3) and B (1, -1):
The slope \( m \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{1 - 1} \]
Since the slope is undefined (vertical line), the equation of line AB is:
\[ x = 1 \]

**Equation of line AC:**
Using points A (1, 3) and C (k, 1):
The slope \( m \) is given by:
\[ m = \frac{1 - 3}{k - 1} = \frac{-2}{k - 1} \]
Using point-slope form:
\[ y - 3 = \frac{-2}{k - 1}(x - 1) \]

Since we found \( k = 5 \) from the previous calculations, substituting \( k \):
\[ y - 3 = \frac{-2}{5 - 1}(x - 1) \]
\[ y - 3 = \frac{-2}{4}(x - 1) \]
\[ y - 3 = -\frac{1}{2}(x - 1) \]
\[ 2(y - 3) = -(x - 1) \]
\[ 2y - 6 = -x + 1 \]
\[ x + 2y - 7 = 0 \]

### Final Results:
- Coordinates of B: (1, -1)
- Coordinates of C: (5, 1)
- Centroid G: (1, 1)
- Equation of line AB: \( x = 1 \)
- Equation of line AC: \( x + 2y - 7 = 0 \)