the parallelogram PRQS has vertices P=(7,12) R=(20,5) and S= (4,3) find the cooridnates of point Q and angle PSR

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.

The parallelogram PRQS has vertices P=(7,12), R=(20,5), and S=(4,3). Find the coordinates of point Q and the angle PSR.

Learn how to find the coordinates of point Q and calculate the angle PSR in the parallelogram PRQS with vertices P(7,12), R(20,5), and S(4,3). This problem covers key geometry concepts, including the midpoint formula, dot product, and the properties of parallelograms. Suitable for students in Grades 10-12.

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