(c) \( \mathrm{A}(-2 ; 3), \mathrm{B}(x ; y), \mathrm{C}(1 ; 4) \) and \( \mathrm{D}(-1 ; 2) \) are the vertices of parallelogram ABCD . Find \( \mathrm{B}(x ; y) \).

To find the coordinates of point B, we can use the property of the diagonals of a parallelogram: the diagonals bisect each other. Thus, the midpoint of diagonal AC must equal the midpoint of diagonal BD. 1. **Calculate the midpoint of AC:** \[ A(-2, 3) \quad \text{and} \quad C(1, 4) \] Midpoint \( M_{AC} = \left( \frac{-2 + 1}{2}, \frac{3 + 4}{2} \right) = \left( -\frac{1}{2}, \frac{7}{2} \right) \) 2. **Let the coordinates of point B be \( B(x, y) \) and point D be \( D(-1, 2) \). Now, we calculate the midpoint of BD:** \[ M_{BD} = \left( \frac{x - 1}{2}, \frac{y + 2}{2} \right) \] 3. **Set the midpoints equal:** \[ \left( -\frac{1}{2}, \frac{7}{2} \right) = \left( \frac{x - 1}{2}, \frac{y + 2}{2} \right) \] 4. **Solve for x and y:** - For the x-coordinates: \[ -\frac{1}{2} = \frac{x - 1}{2} \implies -1 = x - 1 \implies x = 0 \] - For the y-coordinates: \[ \frac{7}{2} = \frac{y + 2}{2} \implies 7 = y + 2 \implies y = 5 \] So, the coordinates of point B are \( B(0, 5) \).