A rectangle \( A B C D \) is to be drawn on a centimetre grid such that \( A \) has coordinates \( (-4,-2) \) \( B \) has coordinates \( (1,10) \) \( C \) has coordinates \( (19, a) \) \( D \) has coordinates \( (b, c) \) Work out the value of \( a \), the value of \( b \) and the value of \( c \). (4 marks) \( \theta a= \) \( \theta b= \) \( \theta c= \)

To determine the values of \( a \), \( b \), and \( c \) for rectangle \( ABCD \) with the given coordinates: - \( A = (-4, -2) \) - \( B = (1, 10) \) - \( C = (19, a) \) - \( D = (b, c) \) **Step 1: Determine \( b \) and \( c \)** Since \( ABCD \) is a rectangle, the diagonals \( AC \) and \( BD \) must bisect each other. Therefore, the midpoints of \( AC \) and \( BD \) must be equal. - **Midpoint of \( AC \):** \[ \left( \frac{-4 + 19}{2}, \frac{-2 + a}{2} \right) = \left( \frac{15}{2}, \frac{a - 2}{2} \right) \] - **Midpoint of \( BD \):** \[ \left( \frac{1 + b}{2}, \frac{10 + c}{2} \right) \] Setting the midpoints equal: \[ \frac{1 + b}{2} = \frac{15}{2} \quad \Rightarrow \quad b = 14 \] \[ \frac{10 + c}{2} = \frac{a - 2}{2} \quad \Rightarrow \quad c = a - 12 \] **Step 2: Determine \( a \)** For \( ABCD \) to be a rectangle, adjacent sides must be perpendicular. Compute the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \): - \( \overrightarrow{AB} = (1 - (-4), 10 - (-2)) = (5, 12) \) - \( \overrightarrow{AD} = (14 - (-4), c - (-2)) = (18, a - 10) \) The dot product of these vectors must be zero: \[ 5 \cdot 18 + 12 \cdot (a - 10) = 0 \quad \Rightarrow \quad 90 + 12a - 120 = 0 \quad \Rightarrow \quad 12a = 30 \quad \Rightarrow \quad a = \frac{30}{12} = \frac{5}{2} \] **Step 3: Substitute \( a \) to find \( c \)** \[ c = a - 12 = \frac{5}{2} - 12 = \frac{5 - 24}{2} = \frac{-19}{2} \] **Final Values:** \[ \theta a = \frac{5}{2} \] \[ \theta b = 14 \] \[ \theta c = \frac{-19}{2} \] **Answer Summary:** \[ \theta a = \frac{5}{2}, \quad \theta b = 14, \quad \theta c = \frac{-19}{2} \]