A rhombus has one vertex at point P(−1, 2) and its diagonals intersect at the origin. If the length of one diagonal is 6 units, what are the possible coordinates for the other three vertices?

To determine the possible coordinates for the vertices of the rhombus, let's follow these steps: 1. **Understanding the Rhombus Properties:** - A rhombus has diagonals that **bisect each other at right angles**. - If one vertex is given and the diagonals intersect at the origin, the opposite vertex is the reflection of the given vertex across the origin. 2. **Given Information:** - One vertex is \( P(-1, 2) \). - The diagonals intersect at the origin \( O(0, 0) \). - The length of one diagonal is 6 units. 3. **Finding the Opposite Vertex:** - Since the diagonals bisect each other at the origin, the opposite vertex \( Q \) of \( P \) is \( Q(1, -2) \). 4. **Determining the Other Diagonal:** - The diagonal \( PQ \) has a length of: \[ \text{Length of } PQ = \sqrt{(1 - (-1))^2 + (-2 - 2)^2} = \sqrt{4 + 16} = 2\sqrt{5} \] Since 6 units is longer than \( 2\sqrt{5} \) (~4.472), the other diagonal must be of length 6 units. 5. **Calculating the Remaining Vertices:** - Let the remaining vertices be \( R(a, b) \) and \( S(-a, -b) \). - The length of diagonal \( RS \) is 6 units: \[ 2\sqrt{a^2 + b^2} = 6 \quad \Rightarrow \quad \sqrt{a^2 + b^2} = 3 \quad \Rightarrow \quad a^2 + b^2 = 9 \] - Since diagonals are perpendicular, the vector for \( RS \) must be perpendicular to the vector \( PQ \) which is \( (2, -4) \). Thus, \( RS \) can have direction vectors proportional to \( (4, 2) \) or \( (-4, -2) \). 6. **Final Coordinates:** - Scaling the direction vectors to have a length of 3 units: \[ R = \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}\right) \quad \text{and} \quad S = \left(-\frac{6\sqrt{5}}{5}, -\frac{3\sqrt{5}}{5}\right) \] Therefore, the possible coordinates for the other three vertices of the rhombus are: - **Vertex \( Q \):** \( (1, -2) \) - **Vertices \( R \) and \( S \):** \( \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}\right) \) and \( \left(-\frac{6\sqrt{5}}{5}, -\frac{3\sqrt{5}}{5}\right) \) **Summary of Possible Coordinates:** \[ \boxed{ \begin{aligned} & (1,\ -2), \\ & \left(\dfrac{6\,\sqrt{5}}{5},\ \dfrac{3\,\sqrt{5}}{5}\right), \\ & \left(-\dfrac{6\,\sqrt{5}}{5},\ -\dfrac{3\,\sqrt{5}}{5}\right) \end{aligned} } \]