求这题的解,要有详细过程
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A(0, 2), C(4, 0)
按题意, 直线在y轴上的截距b > 2
令y = -x/2 + b = 2, x = 2(b - 2), M(2(b - 2), 2)
令直线中x = 4, y = b - 2, N(4, b - 2)
又令直线中y = 0, x = 2b, 直线与x轴的交点为P(2b, 0)
△OMN的面积 = △OMP的面积 - △ONP的面积 = (1/2)OP*M的纵坐标 - (1/2)OP*N的纵坐标
= (1/2)*OP(M的纵坐标 - N的纵坐标)
= (1/2)*2b*[2 - (b - 2)] = b(4 - b) = 3
(b - 1)(b - 3)=0
b = 3, y = -x/2 + 3
舍去b = 1 < 2
按题意, 直线在y轴上的截距b > 2
令y = -x/2 + b = 2, x = 2(b - 2), M(2(b - 2), 2)
令直线中x = 4, y = b - 2, N(4, b - 2)
又令直线中y = 0, x = 2b, 直线与x轴的交点为P(2b, 0)
△OMN的面积 = △OMP的面积 - △ONP的面积 = (1/2)OP*M的纵坐标 - (1/2)OP*N的纵坐标
= (1/2)*OP(M的纵坐标 - N的纵坐标)
= (1/2)*2b*[2 - (b - 2)] = b(4 - b) = 3
(b - 1)(b - 3)=0
b = 3, y = -x/2 + 3
舍去b = 1 < 2
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