Given the points \( A(3 ; 5), B(0 ;-1), C(-8 ; 3) \) and \( D(-4 ; n) \), solve for a) \( A B \) is parallel to \( C D \) b) \( A B \) is perpendicular to \( C D \) \( D \) is the midpoint of \( B C \).

Let's analyze each part of the problem step by step. ### Given Points: - \( A(3, 5) \) - \( B(0, -1) \) - \( C(-8, 3) \) - \( D(-4, n) \) ### a) \( AB \) is parallel to \( CD \) **Vectors:** - **Vector \( \overrightarrow{AB} \):** \[ \overrightarrow{AB} = B - A = (0 - 3, -1 - 5) = (-3, -6) \] - **Vector \( \overrightarrow{CD} \):** \[ \overrightarrow{CD} = D - C = (-4 - (-8), n - 3) = (4, n - 3) \] **Condition for Parallelism:** Two vectors are parallel if one is a scalar multiple of the other. So, \[ \overrightarrow{AB} = k \cdot \overrightarrow{CD} \] \[ (-3, -6) = k \cdot (4, n - 3) \] This gives us the system of equations: \[ -3 = 4k \quad \text{and} \quad -6 = k(n - 3) \] **Solving for \( k \):** \[ k = \frac{-3}{4} \] **Substitute \( k \) into the second equation:** \[ -6 = \left(\frac{-3}{4}\right)(n - 3) \] \[ -6 = \frac{-3}{4}n + \frac{9}{4} \] \[ -6 - \frac{9}{4} = \frac{-3}{4}n \] \[ -\frac{33}{4} = \frac{-3}{4}n \] \[ n = 11 \] **Answer for part a:** \( n = 11 \) --- ### b) \( AB \) is perpendicular to \( CD \) **Condition for Perpendicularity:** Two vectors are perpendicular if their dot product is zero. \[ \overrightarrow{AB} \cdot \overrightarrow{CD} = 0 \] \[ (-3)(4) + (-6)(n - 3) = 0 \] \[ -12 - 6n + 18 = 0 \] \[ 6 - 6n = 0 \] \[ 6n = 6 \] \[ n = 1 \] **Answer for part b:** \( n = 1 \) --- ### c) \( D \) is the midpoint of \( BC \) **Midpoint Formula:** The midpoint \( D \) of points \( B(x_1, y_1) \) and \( C(x_2, y_2) \) is: \[ D\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] **Applying to Points \( B(0, -1) \) and \( C(-8, 3) \):** \[ D\left( \frac{0 + (-8)}{2}, \frac{-1 + 3}{2} \right) = D(-4, 1) \] Since \( D(-4, n) \), we have: \[ n = 1 \] **Answer for part c:** \( n = 1 \) ### Summary of Solutions: - **a)** \( n = 11 \) - **b)** \( n = 1 \) - **c)** \( n = 1 \)