PQRS is a kite.
Diagonal RQ has equation y = 3x+1 and point P has coordinates (-5, 1).
Note that the diagram is not to scale.
Find the equation of diagonal PS.

To find the equation of diagonal \( PS \) in the kite \( PQRS \), we need to utilize the properties of kites. A kite has two pairs of adjacent sides that are equal, and more importantly for this problem, its diagonals are perpendicular to each other. 

### Key Information:
- Diagonal \( RQ \) has the equation: \( y = 3x + 1 \).
- Point \( P \) has coordinates \( (-5, 1) \).
- Diagonals of a kite are perpendicular to each other.

### Step-by-Step Solution:

1. **Slope of diagonal \( RQ \):**
   The equation \( y = 3x + 1 \) is in slope-intercept form \( y = mx + b \), where \( m \) is the slope.
   - The slope of diagonal \( RQ \) is \( 3 \).

2. **Slope of diagonal \( PS \):**
   Since diagonals of a kite are perpendicular, the slope of diagonal \( PS \) must be the negative reciprocal of the slope of diagonal \( RQ \). 
   - The negative reciprocal of \( 3 \) is \( -\frac{1}{3} \).

3. **Equation of diagonal \( PS \):**
   We now have the slope of diagonal \( PS \) (\( -\frac{1}{3} \)) and a point on it, \( P(-5, 1) \). We use the point-slope form of a line equation:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line.
   Substituting \( m = -\frac{1}{3} \), \( x_1 = -5 \), and \( y_1 = 1 \), we get:
   \[
   y - 1 = -\frac{1}{3}(x + 5)
   \]
   Simplify the equation:
   \[
   y - 1 = -\frac{1}{3}x - \frac{5}{3}
   \]
   Add \( 1 \) to both sides:
   \[
   y = -\frac{1}{3}x - \frac{5}{3} + 1
   \]
   Convert \( 1 \) to a fraction with denominator 3:
   \[
   y = -\frac{1}{3}x - \frac{5}{3} + \frac{3}{3}
   \]
   Simplify:
   \[
   y = -\frac{1}{3}x - \frac{2}{3}
   \]

Thus, the equation of diagonal \( PS \) is:
\[
y = -\frac{1}{3}x - \frac{2}{3}
\]

Would you like more details on any part of this? Let me know if you have any further questions.

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Here are 5 related questions:
1. How do you find the equation of a line given two points?
2. How can you determine if two lines are perpendicular or parallel?
3. What are the properties of the diagonals of different quadrilaterals?
4. How would you find the intersection point of two diagonals in a kite?
5. Can you explain the geometric significance of negative reciprocal slopes?

**Tip:** Always check if two lines are perpendicular by confirming that the product of their slopes is \( -1 \).

PQRS is a kite. Diagonal RQ has equation y = 3x+1 and point P has coordinates (-5, 1). Note that the diagram is not to scale. Find the equation of diagonal PS.

Learn how to find the equation of diagonal PS in the kite PQRS, where diagonal RQ has the equation y = 3x+1 and point P is located at (-5, 1). This solution covers key geometric concepts, such as perpendicular diagonals in kites and the application of point-slope form. Suitable for students in Grades 9-11.

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