To find the coordinates of point $Q$ and the angle $\angle PSR$ in parallelogram $PRQS$, let's go through the solution step by step.

1. Coordinates of Point $Q$

A property of parallelograms is that the diagonals bisect each other. In other words, the midpoint of diagonal $PR$ is the same as the midpoint of diagonal $QS$.

• Midpoint of diagonal $PR$: $M_{PR} = \left( \frac{7 + 20}{2}, \frac{12 + 5}{2} \right) = \left( \frac{27}{2}, \frac{17}{2} \right) = (13.5, 8.5)$

• Midpoint of diagonal $QS$: $M_{QS} = \left( \frac{Q_x + 4}{2}, \frac{Q_y + 3}{2} \right)$ Since $M_{QS} = M_{PR} = (13.5, 8.5)$, we can set the coordinates equal and solve for $Q_x$ and $Q_y$: $\frac{Q_x + 4}{2} = 13.5 \quad \text{and} \quad \frac{Q_y + 3}{2} = 8.5$ Solving these: $Q_x + 4 = 27 \quad \Rightarrow \quad Q_x = 23$ $Q_y + 3 = 17 \quad \Rightarrow \quad Q_y = 14$ Thus, the coordinates of $Q$ are: $Q = (23, 14)$

2. Angle $\angle PSR$

Now, let's calculate the angle $\angle PSR$. This requires the direction vectors $\overrightarrow{PS}$ and $\overrightarrow{RS}$, and we can use the dot product formula to find the angle between these vectors.

• Vector $\overrightarrow{PS} = S - P = (4 - 7, 3 - 12) = (-3, -9)$
• Vector $\overrightarrow{RS} = S - R = (4 - 20, 3 - 5) = (-16, -2)$

The dot product of these two vectors is: $\overrightarrow{PS} \cdot \overrightarrow{RS} = (-3)(-16) + (-9)(-2) = 48 + 18 = 66$ The magnitudes of the vectors are: $|\overrightarrow{PS}| = \sqrt{(-3)^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90}$ $|\overrightarrow{RS}| = \sqrt{(-16)^2 + (-2)^2} = \sqrt{256 + 4} = \sqrt{260}$ Using the dot product formula for the angle between two vectors: $\cos \theta = \frac{\overrightarrow{PS} \cdot \overrightarrow{RS}}{|\overrightarrow{PS}| |\overrightarrow{RS}|} = \frac{66}{\sqrt{90} \times \sqrt{260}}$ Calculating the cosine: $\cos \theta = \frac{66}{\sqrt{90 \times 260}} = \frac{66}{\sqrt{23400}} = \frac{66}{153}$ Thus, $\theta = \cos^{-1} \left( \frac{66}{153} \right) \approx \cos^{-1}(0.431)$ Using an inverse cosine function: $\theta \approx 64.44^\circ$

Therefore, the angle $\angle PSR$ is approximately $64.44^\circ$.

Summary:

• The coordinates of $Q$ are $(23, 14)$.
• The angle $\angle PSR$ is approximately $64.44^\circ$.

Would you like further details or have any questions?

Related Questions:

1. How do you compute the area of a parallelogram given its vertices?
2. What are the properties of diagonals in a parallelogram?
3. How can you find the other angles in a parallelogram?
4. How can the coordinates of a missing vertex in a parallelogram be found geometrically?
5. What are some real-world applications of parallelogram properties in physics or engineering?

Tip:

Always remember that diagonals of a parallelogram bisect each other, which is a key property for finding missing points or solving geometry problems.