Question

In triangle $\triangle ABC$, $A=(1,2)$ and the equations of medians through $B$ and $C$ are $x+y=5$ and $x=4$. Find area of triangle ABC.

• A. $12$
• B. $9$
• C. $4$
• D. $5$

Hint:

Centroid and Area: Use centroid formulas from medians’ intersection, then plug back into vertex expressions. Use determinant or Shoelace formula to find area.

Answer 1

The Correct Option is B

The centroid $G$ lies at the intersection of the medians: \[ x+y = 5, \quad x = 4 \Rightarrow y = 1 \Rightarrow G = (4,1) \] Let $A = (1,2)$, $G = \left( \frac{1+x_B+x_C}{3}, \frac{2+y_B+y_C}{3} \right) = (4,1)$. Solving: \[ x_B + x_C = 11, \quad y_B + y_C = 1 \] Given $B$ lies on $x+y=5$: $x_B + y_B = 5$ and $C$ lies on $x=4$: $x_C = 4$ \[ x_B = 7, \; y_B = -2, \; y_C = 3 \Rightarrow B = (7, -2), \; C = (4, 3) \] Using determinant formula: \[ \text{Area} = \frac{1}{2} |1(-2 - 3) + 7(3 - 2) + 4(2 + 2)| = \frac{1}{2} | -5 + 7 + 16 | = \frac{18}{2} = 9 \]