To find the area of triangle ABC given the coordinates of point A and the equations of the medians through points B and C, we can follow these steps:
### Step 1: Identify the coordinates of point A
The coordinates of point A are given as \( A(1, 2) \).
### Step 2: Determine the equations of the medians
The equations of the medians are given as:
- Median through B: \( x + y = 5 \)
- Median through C: \( x = 4 \)
### Step 3: Find the coordinates of point B
Let the coordinates of point B be \( B(p, q) \). Since point B lies on the median \( x + y = 5 \), we can express this as:
\[ p + q = 5 \]
This is our first equation.
### Step 4: Find the coordinates of point C
Let the coordinates of point C be \( C(r, s) \). Since point C lies on the median \( x = 4 \), we have:
\[ r = 4 \]
This gives us the x-coordinate of point C.
### Step 5: Find the midpoint D of segment AB
The midpoint D of segment AB can be expressed as:
\[ D\left(\frac{1 + p}{2}, \frac{2 + q}{2}\right) \]
### Step 6: Find the coordinates of point D using the median through C
Since D is also the midpoint of AC, we can express the coordinates of D using the coordinates of C:
\[ D\left(\frac{1 + 4}{2}, \frac{2 + s}{2}\right) = D\left(\frac{5}{2}, \frac{2 + s}{2}\right) \]
### Step 7: Set the two expressions for D equal to each other
Equating the two expressions for D:
\[
\frac{1 + p}{2} = \frac{5}{2} \quad \text{and} \quad \frac{2 + q}{2} = \frac{2 + s}{2}
\]
From the first equation:
\[
1 + p = 5 \implies p = 4
\]
From the second equation:
\[
2 + q = 2 + s \implies q = s
\]
### Step 8: Substitute p into the equation for B
Now substituting \( p = 4 \) into the equation \( p + q = 5 \):
\[
4 + q = 5 \implies q = 1
\]
Thus, the coordinates of point B are \( B(4, 1) \).
### Step 9: Find the coordinates of point C
We already found that \( r = 4 \), and since \( q = s \), we have:
\[
C(4, 1)
\]
### Step 10: Calculate the area of triangle ABC
Now we have the coordinates:
- \( A(1, 2) \)
- \( B(4, 1) \)
- \( C(4, 3) \)
Using the formula for the area of a triangle given vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
\text{Area} = \frac{1}{2} \left| 1(1 - 3) + 4(3 - 2) + 4(2 - 1) \right|
\]
Calculating:
\[
= \frac{1}{2} \left| 1(-2) + 4(1) + 4(1) \right|
= \frac{1}{2} \left| -2 + 4 + 4 \right|
= \frac{1}{2} \left| 6 \right|
= 3
\]
### Final Area Calculation
Thus, the area of triangle ABC is \( 3 \) square units.
