Question

In a quadrilateral PQRS, M and N are mid-points of the sides PQ and RS respectively. If $\overline{PS} + \overline{QR} = t\overline{MN}$, then $t =$

• A. $\frac{1}{2}$
• B. $4$
• C. $\frac{3}{2}$
• D. $2$

Hint:

This is a standard vector geometry property: In any quadrilateral, the vector sum of two opposite sides is always equal to exactly twice the vector connecting the midpoints of the other two sides.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an arbitrary quadrilateral with midpoints defined on two opposite sides. We need to find the scalar multiplier $t$ that relates the sum of two specific side vectors to the segment connecting the midpoints.

Step 2: Key Formula or Approach:
Use position vectors to represent the vertices. Let the position vectors of the vertices $P, Q, R, S$ with respect to an arbitrary origin be $\vec{p}, \vec{q}, \vec{r}, \vec{s}$.
The vector representing a directed line segment from $A$ to $B$ is $\overline{AB} = \vec{b} - \vec{a}$.
The midpoint formula for a position vector is $\vec{m} = \frac{\vec{a} + \vec{b}}{2}$.

Step 3: Detailed Explanation:
First, find the position vectors of the midpoints $M$ and $N$:
Since $M$ is the midpoint of $PQ$, $\vec{m} = \frac{\vec{p} + \vec{q}}{2}$.
Since $N$ is the midpoint of $RS$, $\vec{n} = \frac{\vec{r} + \vec{s}}{2}$.
Next, express the vector $\overline{MN}$ in terms of the position vectors:
$$\overline{MN} = \vec{n} - \vec{m} = \left(\frac{\vec{r} + \vec{s}}{2}\right) - \left(\frac{\vec{p} + \vec{q}}{2}\right)$$ $$\overline{MN} = \frac{\vec{r} + \vec{s} - \vec{p} - \vec{q}}{2}$$ Multiply both sides by 2 to clear the fraction:
$$2\overline{MN} = \vec{r} + \vec{s} - \vec{p} - \vec{q}$$ Now, let's rearrange the right side to group the vectors that correspond to the sides $PS$ and $QR$:
Group $\vec{s}$ and $-\vec{p}$ together: $(\vec{s} - \vec{p}) = \overline{PS}$
Group $\vec{r}$ and $-\vec{q}$ together: $(\vec{r} - \vec{q}) = \overline{QR}$
Substitute these segment vectors back into the equation:
$$2\overline{MN} = (\vec{s} - \vec{p}) + (\vec{r} - \vec{q})$$ $$2\overline{MN} = \overline{PS} + \overline{QR}$$ The problem states that $\overline{PS} + \overline{QR} = t\overline{MN}$.
By direct comparison, $t = 2$.

Step 4: Final Answer:
The value of $t$ is $2$, which corresponds to option (D).