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2. \( A B C D \) is a parallelogram such that \( A(5,2) \), \( B(-2,-3) \) and \( \overrightarrow{A C}=\binom{1}{-3} \) Find (i) the coordinates of \( C \) and \( D \); (ii) \( \overrightarrow{B C} \).

Ask by Bowen Hamilton. in Ghana
Jan 23,2025

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**(i)** Point \( C \) is at \( (6, -1) \) and point \( D \) is at \( (13, 4) \). **(ii)** The vector \( \overrightarrow{BC} \) is \( \binom{8}{2} \).

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Mind Expander

To find the coordinates of point \( C \) and point \( D \) in the parallelogram \( ABCD \), we first calculate the coordinates of \( C \). Given that \( \overrightarrow{AC} = \binom{1}{-3} \), we can find \( C \) starting from point \( A(5, 2) \): 1. \( C = A + \overrightarrow{AC} = (5, 2) + (1, -3) = (5 + 1, 2 - 3) = (6, -1) \). Next, we use the property of parallelograms, where the midpoints of diagonals \( AC \) and \( BD \) are the same. Therefore, the coordinates of point \( D \) can be found as follows: 2. The coordinates of point \( B(-2, -3) \) give the equation for midpoint: \[ \text{Midpoint of } AC = \left( \frac{5 + 6}{2}, \frac{2 + (-1)}{2} \right) = \left( \frac{11}{2}, \frac{1}{2} \right) \] and the midpoint of \( BD \) is: \[ \text{Midpoint of } BD = \left( \frac{-2 + x_D}{2}, \frac{-3 + y_D}{2} \right), \] where \( D(x_D, y_D) \). 3. Setting these equal gives us two equations: \[ \frac{-2 + x_D}{2} = \frac{11}{2} \implies -2 + x_D = 11 \implies x_D = 13, \] \[ \frac{-3 + y_D}{2} = \frac{1}{2} \implies -3 + y_D = 1 \implies y_D = 4. \] Thus, the coordinates of point \( D \) are \( (13, 4) \). Now, let's find \(\overrightarrow{BC}\): Given the coordinates of points \( B(-2, -3) \) and \( C(6, -1) \): 4. \( \overrightarrow{BC} = C - B = (6, -1) - (-2, -3) = (6 + 2, -1 + 3) = (8, 2) \). So, the final answers are: - The coordinates of \( C \) are \( (6, -1) \) and \( D \) are \( (13, 4) \). - The vector \( \overrightarrow{BC} \) is \( \binom{8}{2} \).

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