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: \[ \text{Area} = \frac{1}{2} |1(-2 - 3) + 7(3 - 2) + 4(2 + 2)| = \frac{1}{2} | -5 + 7 + 16 | = \frac{18}{2} = 9 \]

Answer 2

Step 1: Identify given points and medians
We have point \( A = (1,2) \). The median through \( B \) lies on the line \( x + y = 5 \), and the median through \( C \) lies on the line \( x = 4 \).

Step 2: Understand medians and centroid
The medians of a triangle intersect at the centroid \( G \), which divides each median in the ratio 2:1.
If the vertices are \( A, B, C \), and \( M \), \( N \) are midpoints of opposite sides, then the median from \( B \) passes through the midpoint of \( AC \), and the median from \( C \) passes through the midpoint of \( AB \).

Step 3: Calculate midpoints
Let \( B = (x_B, y_B) \) and \( C = (x_C, y_C) \).
- Midpoint of \( AC \) is \( M = \left(\frac{1 + x_C}{2}, \frac{2 + y_C}{2}\right) \).
- Midpoint of \( AB \) is \( N = \left(\frac{1 + x_B}{2}, \frac{2 + y_B}{2}\right) \).

Step 4: Coordinates of centroid \( G \)
\[ G = \left(\frac{1 + x_B + x_C}{3}, \frac{2 + y_B + y_C}{3}\right) \]

Step 5: Median through \( B \) lies on \( x + y = 5 \)
The median through \( B \) connects \( B \) and \( M \). The centroid lies on this median, so \( G \) lies on the line \( x + y = 5 \):
\[ \frac{1 + x_B + x_C}{3} + \frac{2 + y_B + y_C}{3} = 5 \implies (1 + x_B + x_C) + (2 + y_B + y_C) = 15 \]
\[ x_B + x_C + y_B + y_C = 12 \quad \quad (1) \]

Step 6: Median through \( C \) lies on \( x = 4 \)
The median through \( C \) connects \( C \) and \( N \). The centroid \( G \) lies on this median, so \( G \) lies on \( x = 4 \):
\[ \frac{1 + x_B + x_C}{3} = 4 \implies 1 + x_B + x_C = 12 \implies x_B + x_C = 11 \quad \quad (2) \]

Step 7: Use (1) and (2) to find \( y_B + y_C \)
From (1): \( x_B + x_C + y_B + y_C = 12 \), and (2): \( x_B + x_C = 11 \)
So,
\[ 11 + y_B + y_C = 12 \implies y_B + y_C = 1 \quad \quad (3) \]

Step 8: Use median through \( C \) on \( x=4 \) to find midpoint \( N \)
Midpoint \( N = \left(\frac{1 + x_B}{2}, \frac{2 + y_B}{2}\right) \) lies on median through \( C \) which is \( x=4 \), so \( N_x = 4 \):
\[ \frac{1 + x_B}{2} = 4 \implies 1 + x_B = 8 \implies x_B = 7 \]

Step 9: Use (2) to find \( x_C \)
From (2), \( x_B + x_C = 11 \), so
\[ 7 + x_C = 11 \implies x_C = 4 \]

Step 10: Find \( y_B \) and \( y_C \)
We have \( y_B + y_C = 1 \) from (3). Use the median through \( B \) lying on \( x + y = 5 \) to get more info.
The midpoint \( M \) of \( AC \) is:
\[ M = \left(\frac{1 + x_C}{2}, \frac{2 + y_C}{2}\right) = \left(\frac{1 + 4}{2}, \frac{2 + y_C}{2}\right) = (2.5, \frac{2 + y_C}{2}) \]
The median through \( B \) connects \( B(7,y_B) \) and \( M(2.5, \frac{2 + y_C}{2}) \). The slope of the median is:
\[ m = \frac{\frac{2 + y_C}{2} - y_B}{2.5 - 7} = \frac{\frac{2 + y_C}{2} - y_B}{-4.5} \]
This median lies on the line \( x + y = 5 \), or \( y = 5 - x \), which has slope -1.
So,
\[ \frac{\frac{2 + y_C}{2} - y_B}{-4.5} = -1 \implies \frac{2 + y_C}{2} - y_B = 4.5 \]
Multiply both sides by 2:
\[ 2 + y_C - 2 y_B = 9 \implies y_C - 2 y_B = 7 \quad \quad (4) \]

Step 11: Solve system of equations for \( y_B \) and \( y_C \)
From (3): \( y_B + y_C = 1 \), so \( y_C = 1 - y_B \). Substitute into (4):
\[ (1 - y_B) - 2 y_B = 7 \implies 1 - y_B - 2 y_B = 7 \implies 1 - 3 y_B = 7 \]
\[ -3 y_B = 6 \implies y_B = -2 \]
Then
\[ y_C = 1 - (-2) = 3 \]

Step 12: Final vertices of the triangle
\[ A = (1,2), \quad B = (7, -2), \quad C = (4, 3) \]

Step 13: Calculate area of \(\triangle ABC\)
Area formula using coordinates:
\[ \text{Area} = \frac{1}{2} \left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right| \]
Substitute:
\[ = \frac{1}{2} | 1(-2 - 3) + 7(3 - 2) + 4(2 - (-2)) | = \frac{1}{2} | 1(-5) + 7(1) + 4(4) | \]
\[ = \frac{1}{2} | -5 + 7 + 16 | = \frac{1}{2} | 18 | = 9 \]

Final answer: The area of the triangle \( ABC \) is 9.