5. The diagonals of quadrilateral \( E F G H \) intersect at \( D(- \) 1,4). Two vertices of the quadrilateral are \( E(2,7) \) and \( F(-3,5) \). What must be the coordinates of \( G \) and \( H \) to ensure that \( E F G H \) is a parallelogram?

To ensure that quadrilateral \( EFGH \) is a parallelogram, the diagonals must bisect each other. Given that point \( D(-1, 4) \) is the midpoint of both diagonals, we can use this to find the coordinates of \( G \) and \( H \). Since \( D \) is the midpoint of \( EH \), we can set up the equation: \[ D = \left( \frac{E_x + H_x}{2}, \frac{E_y + H_y}{2} \right) \] Substituting in the coordinates for \( E \): \[ (-1, 4) = \left( \frac{2 + H_x}{2}, \frac{7 + H_y}{2} \right) \] From here, solving for \( H_x \) and \( H_y \): 1. \( -1 = \frac{2 + H_x}{2} \) leads to \( H_x = -4 \) 2. \( 4 = \frac{7 + H_y}{2} \) leads to \( H_y = 1 \) So, \( H \) has coordinates \( (-4, 1) \). Now, for \( G \), we also have that \( D \) is the midpoint of \( F \) and \( G \): \[ (-1, 4) = \left( \frac{-3 + G_x}{2}, \frac{5 + G_y}{2} \right) \] Solving for \( G_x \) and \( G_y \): 1. \( -1 = \frac{-3 + G_x}{2} \) gives \( G_x = 1 \) 2. \( 4 = \frac{5 + G_y}{2} \) gives \( G_y = 3 \) Thus, the coordinates for \( G \) are \( (1, 3) \). So, the coordinates of the vertices that ensure \( EFGH \) forms a parallelogram are \( G(1, 3) \) and \( H(-4, 1) \).