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 from \(A(1, 2)\) intersects the midpoint of side \(BC\), denoted as \((x, y)\).

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

Step 1: Determine \(x\). From the median through \(C\):

\[ x = 4. \]

Step 2: Determine \(y\). Substitute \(x = 4\) into the median through \(B\):

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

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

\[ (4, 1). \]

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

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

Final Answer: The midpoint of \(BC\) is:

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