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Step-by-Step Solution
Step 1: Understand the Problem
We are given a triangle ABC with coordinates of A as (1, 2). Two medians are described by the equations:
1) The median through B: x + y = 5
2) The median through C: x = 4
We need to use these median equations to find the missing coordinates of B and C and then determine the area of triangle ABC.
Step 2: Use the Given Median Through C
The median from vertex C is given by x = 4. This means that the point C must have the form (4, y) for some unknown y. Let us denote C as (4, y).
Step 3: Locate the Midpoint of Segment AC
The midpoint D of segment AC should lie on the median through B, which is x + y = 5. Since A is (1, 2) and C is (4, y), the midpoint D is:
$D = \\left( \\frac{1 + 4}{2}, \\frac{2 + y}{2} \\right) = \\left( \\frac{5}{2}, \\frac{2 + y}{2} \\right).$
This midpoint D must satisfy the line x + y = 5.
Step 4: Impose the Condition from the Median Through B
Substitute the coordinates of D into the equation x + y = 5:
$\\left(\\frac{5}{2}\\right) + \\left(\\frac{2 + y}{2}\\right) = 5.
$
Combine the terms on the left side:
$\\frac{5 + (2 + y)}{2} = 5
\\quad \\Longrightarrow \\quad
\\frac{7 + y}{2} = 5
\\quad \\Longrightarrow \\quad
7 + y = 10
\\quad \\Longrightarrow \\quad
y = 3.$
Hence, the coordinates of C are (4, 3).
Step 5: Confirm the Intersection of the Medians
Given the median through C is x = 4 and the median through B is x + y = 5, these lines intersect at the centroid G. Plug x = 4 into x + y = 5:
$4 + y = 5
\\Longrightarrow
y = 1.
$
Therefore, the centroid is G(4, 1). The centroid divides each median in the ratio 2:1.
Step 6: Compute the Area of Triangle ABC
The area of triangle ABC is 3 times the area of triangle AGC (a property of the centroid: it divides the triangle into six smaller triangles of equal area).
A = (1, 2)
G = (4, 1)
C = (4, 3)
First, we find the area of $\\triangle AGC$ using the determinant (matrix) formula for the area of a triangle with coordinates $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:
$\\text{Area} = \\frac{1}{2} \\big|\\, x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \\big|.$
Substitute A(1, 2), G(4, 1), and C(4, 3):
$\\text{Area}(AGC) = \\frac{1}{2}\\Big| 1(1 - 3) + 4(3 - 2) + 4(2 - 1) \\Big|.$
$= \\frac{1}{2}\\Big| 1(-2) + 4(1) + 4(1) \\Big|
= \\frac{1}{2}\\Big| -2 + 4 + 4 \\Big|
= \\frac{1}{2}\\times 6
= 3.
$
Hence, the area of $\\triangle ABC$ is:
$\\triangle ABC = 3 \\times \\triangle AGC = 3 \\times 3 = 9.$
Step 7: Final Answer
The area of $\\triangle ABC$ is 9 square units.
