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:

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