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Mathematics
OpenStudy (anonymous):

/_\ DEF has vertices D(-2,4), E(0,-2), and F(6,2). Determine: a) an equation for DG, the median from D to EF b) an equation for FC, the altitude from F to DE c) an equation for AB, the right bisector of DF

7 years ago
OpenStudy (anonymous):

(a) G is midpoint of EF = ((0+6)/2, (-2+2)/2) = (3,0) Equation of line DG is y-4 = m(x+2), where m = slope of DG m = ( y1-y2)/(x1-x2) = (4-0)/(-2-3) = -4/5 DG is the line y - 4 = -(4/5)(x + 2), so 5y - 20 = -4x - 8 DG is 4x + 5y - 12 = 0

7 years ago
OpenStudy (anonymous):

(b) FC perpendicular to DE. Slope of DE = (4--2)/(-2-0) 6/(-2) = -3 so slope of FC = 1/3 (remember m_1 times m_2 = -1) So equation of line FC is y-2 = (1/3)(x-6) 3y - 6 = x - 6 FC is 3y - x = 0

7 years ago
OpenStudy (anonymous):

(c) Mid point of DF = ((-2+6)/2, (4+2)/2) = (2,3) Slope of DF = (4-2)/(-2-6) = -2/8 = -1/4, so slope of right bisector = 4 AB is the line y-3 = 4(x - 2), or y + 4x +5 = 0

7 years ago
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