如图,在平面直角坐标系中,点O为坐标原点,直线y=-3/4x+6交X轴于点A,交y轴于点B,BC与x轴平行,AC为△OA

(1) A(8,0),B(0,6)从做角OAB的平分线,交y轴于D；从D做AB的垂线,交AB于E。△OAD。△EAD全等,OA² = EA²E(e,6 - 3e/4),0 < e < 88² = (e - 8)² + (0 - 6 + 3e/4)²(e - 8)² = 64*16/25e - 8 = ±32/5e = 8/5 (舍去另一解> 8)E(8/5,24/5)DE² = OD²D(0,d)(0 - 8/5)² + (d - 24/5)² = d²d = 8/3D(0,8/3)AD的斜率k = (8/3 - 0)/(0 - 8) = -1/3AC与垂直AC的斜率= -1/k = 3AC的方程:y - 0 = 3(x - 8),y = 3(x - 8)y = 6,x = 10C(10,6)y = k/x,6 = k/10,k = 60原来以为没学过点到直线的距离,否则更直接。(2)BC = 10,需10/2 = 5秒OC = 6,需6/2 = 3秒0 ≤ t ≤ 5:BP = 10 - 2tS = (1/2)BP*OB = (1/2)(10 - 2t)*6 = 30 - 6t5< t ≤ 8:S = (1/2)BP*OA = (1/2)*2(t - 5)*8 = 8(t - 5)(3)AC的方程3x - y - 24 = 0E(e,6 - 3e/4)到直线AC的距离d = |3e - 6 + 3e/4 - 24|/√(3² + 1) = 3√10/2e = 12或e = -4点E不在AB上