Given the points: \( P(-1 ; 4), Q(2 ; 7), R(-3 ; 2) \) and \( S(m ;-1) \), solve for a) \( P Q \) is parallel to \( R S \) b) \( P Q \) is perpendicular to \( R S \).

Let's analyze the problem step by step for both parts. ### Given Points: - \( P(-1, 4) \) - \( Q(2, 7) \) - \( R(-3, 2) \) - \( S(m, -1) \) ### Vectors: 1. **Vector \( \vec{PQ} \):** \[ \vec{PQ} = Q - P = (2 - (-1), 7 - 4) = (3, 3) \] 2. **Vector \( \vec{RS} \):** \[ \vec{RS} = S - R = (m - (-3), -1 - 2) = (m + 3, -3) \] --- ### a) \( \vec{PQ} \) is Parallel to \( \vec{RS} \) For two vectors to be parallel, one must be a scalar multiple of the other. Therefore, there exists a scalar \( k \) such that: \[ \vec{PQ} = k \cdot \vec{RS} \] \[ (3, 3) = k(m + 3, -3) \] This gives us the system of equations: \[ 3 = k(m + 3) \quad \text{(1)} \] \[ 3 = k(-3) \quad \text{(2)} \] From equation (2): \[ 3 = -3k \implies k = -1 \] Substitute \( k = -1 \) into equation (1): \[ 3 = -1 \cdot (m + 3) \implies 3 = -m - 3 \implies m = -6 \] **Answer for part a:** \( m = -6 \) --- ### b) \( \vec{PQ} \) is Perpendicular to \( \vec{RS} \) For two vectors to be perpendicular, their dot product must be zero: \[ \vec{PQ} \cdot \vec{RS} = 0 \] \[ (3)(m + 3) + (3)(-3) = 0 \] \[ 3m + 9 - 9 = 0 \implies 3m = 0 \implies m = 0 \] **Answer for part b:** \( m = 0 \) ### Summary: - **a)** \( m = -6 \) - **b)** \( m = 0 \) Answer: a) \( m = -6 \) b) \( m = 0 \)