Question

In \( \triangle ABC \), coordinates of A are (1, 2). If the equations of the medians through B and C are \( x+y=5 \) and \( x=4 \) respectively, then the area of \( \triangle ABC \) (in sq. units) is

• A. 12
• B. 9
• C. 6
• D. 4

Hint:

Determinant area shortcut: Form relative vectors from \( A \): \( \vec{AB} = (6, -4) \), \( \vec{AC} = (3, 1) \). \[ \text{Area} = \frac{1}{2} |6(1) - (-4)(3)| = \frac{1}{2} |6 + 12| = 9 \] This avoids using the long multi-term coordinate area formula!

Answer 1

The Correct Option is B

Concept: The intersection point of any two medians of a triangle uniquely defines its centroid \( G \). Once the coordinates of vertex \( A \) and centroid \( G \) are known, we can calculate the area of the triangle using the centroid-area relationship or by determining the remaining vertices via properties of medians.

Step 1: Finding the coordinates of the centroid \( G \).
Solving the median lines: \[ x = 4 \quad \text{and} \quad x + y = 5 \implies 4 + y = 5 \implies y = 1 \] Thus, the centroid is \( G(4, 1) \).

Step 2: Determining vertex locations or using vector coordinates.
Let \( B = (x_2, y_2) \) and \( C = (4, y_3) \) since \( C \) lies on the vertical line \( x=4 \). Since \( B \) lies on \( x+y=5 \), we have \( y_2 = 5 - x_2 \). Using the centroid formula \( G = \frac{A+B+C}{3} \): \[ \frac{1 + x_2 + 4}{3} = 4 \implies 5 + x_2 = 12 \implies x_2 = 7 \] This gives \( y_2 = 5 - 7 = -2 \), so \( B = (7, -2) \). Now finding the y-coordinate of \( C \): \[ \frac{2 + (-2) + y_3}{3} = 1 \implies y_3 = 3 \implies C = (4, 3) \]

Step 3: Calculating the triangle area.
Vertices are \( A(1, 2) \), \( B(7, -2) \), and \( C(4, 3) \). \[ \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \] \[ = \frac{1}{2} |1(-2 - 3) + 7(3 - 2) + 4(2 - (-2))| = \frac{1}{2} |-5 + 7 + 16| = \frac{1}{2} |18| = 9 \]