Question

The vertices of a rhombus are located at (a, 0), (0, b), (-a, 0), and (0, -b). What is the midpoint of the side that is in quadrant II?

Answer 1

To find the midpoint of the side of the rhombus located in quadrant II, let's first identify the vertices of the rhombus and the corresponding sides.

The vertices of the rhombus are given as follows:
- \( (a, 0) \)
- \( (0, b) \)
- \( (-a, 0) \)
- \( (0, -b) \)

The pairs of vertices that form the sides of the rhombus are:
1. \( (a, 0) \) and \( (0, b) \)
2. \( (0, b) \) and \( (-a, 0) \)
3. \( (-a, 0) \) and \( (0, -b) \)
4. \( (0, -b) \) and \( (a, 0) \)

We need to determine which of these sides lies in quadrant II. The vertices in quadrant II are those with negative x-coordinates and positive y-coordinates.

From the vertices:
- \( (0, b) \) is in quadrant II.
- \( (-a, 0) \) is on the x-axis (not in any quadrant), and has negative x-coordinates.

However, the side that connects these two vertices is the side between \( (0, b) \) and \( (-a, 0) \).

Now, to find the midpoint of this side, we apply the midpoint formula. If we have two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the midpoint \( M \) is given by:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

Here, \( (x_1, y_1) = (0, b) \) and \( (x_2, y_2) = (-a, 0) \).

Now, we compute the midpoint:
\[
M = \left( \frac{0 + (-a)}{2}, \frac{b + 0}{2} \right) = \left( \frac{-a}{2}, \frac{b}{2} \right)
\]

Thus, the midpoint of the side that is located in quadrant II is:
\[
\boxed{\left( -\frac{a}{2}, \frac{b}{2} \right)}
\]