Question

In \(\triangle ABC\), coordinates of \(A\) are \((1, 2)\), and the equations of the medians through \(B\) and \(C\) are \(x + y = 5\) and \(x = 4\), respectively. Then the midpoint of \(BC\) is:

• A. \(\left(5, \frac{1}{2}\right)\)
• B. \(\left(\frac{11}{2}, 1\right)\)
• C. \(\left(11, \frac{1}{2}\right)\)
• D. \(\left(\frac{11}{2}, \frac{1}{2}\right)\)

Answer 1

The Correct Option is D

The median through \(A(1, 2)\) passes through the midpoint of side \(BC\). Let the midpoint of \(BC\) be \((x, y)\).

1. The median through \(B\) is given by \(x + y = 5\).
2. The median through \(C\) is given by \(x = 4\).

Step 1: Solve for \(x\). From the equation of the median through \(C\), we know:

\[ x = 4. \]

Step 2: Solve for \(y\). Substitute \(x = 4\) into the equation of the median through \(B\):

\[ 4 + y = 5 \implies y = 1. \]

Thus, the midpoint of \(BC\) is:

\[ (4, 1). \]

Step 3: Adjust for the centroid. The centroid divides the median in the ratio \(2 : 1\). Since \(A\) is at \((1, 2)\), the coordinates of the midpoint of \(BC\) are:

\[ \left(\frac{11}{2}, \frac{1}{2}\right). \]

Conclusion: The midpoint of \(BC\) is:

\[ \boxed{\left(\frac{11}{2}, \frac{1}{2}\right)}. \]