Find the coordinates of N, given A(6; -2), B(2; 15), C(-4; 3), and M is the midpoint of AB and MN || BC.

Step 1: Determine the coordinates of M, the midpoint of AB. The midpoint formula for two points (x_1, y_1) and (x_2, y_2) is M = ((x_1+x_2)/(2), (y_1+y_2)/(2)). Given A(6; -2) and B(2; 15). M = ((6+2)/(2), (-2+15)/(2)) M = ((8)/(2), (13)/(2)) M = (4, (13)/(2)) The coordinates of M are (4, (13)/(2)). Step 2: Determine the gradient of line MN. Since MN BC, their gradients are equal. The gradient formula for two points (x_1, y_1) and (x_2, y_2) is m = (y_2-y_1)/(x_2-x_1). Given B(2; 15) and C(-4; 3). m_BC = (3-15)/(-4-2) m_BC = (-12)/(-6) m_BC = 2 Therefore, the gradient of line MN is 2. Step 3: Determine the equation of line MN in the form y = mx+c. We have the gradient m_MN = 2 (from Step 2) and point M(4, (13)/(2)) (from Step 1). Using the point-slope form y - y_1 = m(x - x_1): y - (13)/(2) = 2(x - 4) y - (13)/(2) = 2x - 8 y = 2x - 8 + (13)/(2) y = 2x - (16)/(2) + (13)/(2) y = 2x - (3)/(2) The equation of line MN is y = 2x - (3)/(2). Step 4: Calculate, with reasons, the coordinates of point N. Given that M is the midpoint of AB and MN BC. By the Midpoint Theorem (or its converse), if a line passes through the midpoint of one side of a triangle and is parallel to another side, then it bisects the third side. Therefore, N is the midpoint of AC. Given A(6; -2) and C(-4; 3). N = ((6+(-4))/(2), (-2+3)/(2)) N = ((2)/(2), (1)/(2)) N = (1, (1)/(2)) The coordinates of N are (1, (1)/(2)). Step 5: If ABCD (in that order) is a parallelogram, determine the coordinates of point D. In a parallelogram, the diagonals bisect each other. This means the midpoint of AC is the same as the midpoint of BD. From Step 4, the midpoint of AC is N(1, (1)/(2)). Let D = (x_D, y_D). The midpoint of BD is ((x_B+x_D)/(2), (y_B+y_D)/(2)). Given B(2; 15). Equating the midpoints: (2+x_D)/(2) = 1 2+x_D = 2 x_D = 0 (15+y_D)/(2) = (1)/(2) 15+y_D = 1 y_D = 1-15 y_D = -14 The coordinates of D are (0, -14). Send me the next one 📸