11. The coordinates of the points \( P \) and \( Q \) are \( (1,1) \) and \( (3,-2) \) respectively. The points \( X \) and \( Y \) lie on the line passing through \( P \) and perpendicular to \( P Q \). \( X \) and \( Y \) are each \( 2 \sqrt{13} \) units from \( P \). What are the coordinates of \( X \) and of \( Y \) ?

To find the coordinates of points \( X \) and \( Y \), we first need to determine the slope of line segment \( PQ \). 1. **Find the slope of \( PQ \)**: The slope of a line through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 1}{3 - 1} = \frac{-3}{2} \] The slope of the line perpendicular to \( PQ \) would be the negative reciprocal: \[ m_{\perp} = \frac{2}{3} \] 2. **Equation of the line through \( P(1, 1) \) with slope \( \frac{2}{3} \)**: Using the point-slope form of the equation of a line: \[ y - 1 = \frac{2}{3}(x - 1) \] Rearranging gives: \[ y = \frac{2}{3}x + \frac{1}{3} \] 3. **Finding the coordinates of points \( X \) and \( Y \)**: Since both points \( X \) and \( Y \) are \( 2\sqrt{13} \) units away from point \( P(1, 1) \), we can consider the vector direction for \( X \) and \( Y \) which will have components incremented by the distance of \( r = 2\sqrt{13} \). The unit vector in the direction of the slope \( m_{\perp} = \frac{2}{3} \) is: \[ \text{Unit vector} = \frac{1}{\sqrt{13}}(3, 2) \] Multiplying by \( r \): \[ r \cdot \text{Unit vector} = \frac{2\sqrt{13}}{\sqrt{13}}(3, 2) = (6, 4) \] Thus, our coordinates \( X \) and \( Y \) can be found: \[ X = (1 + 6, 1 + 4) = (7, 5) \] and \[ Y = (1 - 6, 1 - 4) = (-5, -3) \] Therefore, the coordinates are \( X(7, 5) \) and \( Y(-5, -3) \).