Question

在平面直角坐标系中,A(8,0) B(0,3),点C(x,0)是线段OA上一动点(不与端点O、A重合),

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跟知道这道题一样
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Answer 1

(1)有两种可能
(a)∠OBC = ∠ACD
tan∠OBC = tan∠ACD
OC/OB = AD/CA
x/3 = y/(8-x)
即y = (8x-x²)/3

(b) ∠OBC = ∠ADC
tan∠OBC = tan∠ADC
OC/OB = CA/AD
x/3 = (8-x)/y
y = 3(8-x)/x

(2)为避免混淆，设C(c, 0)
BC的斜率为(3-0)/(0-c) = -3/c
CD的斜率为c/3
CD的方程： y - 0 = (c/3)(x - c)
y = c(x-c)/3
取x = 8, y = c(8-c)/3
D(8, c(8-c)/3)
梯形AOBD的面积S = (OB + AD)*OA/2
=(3 + c(8-c)/3)*8/2
= 12 + 4(8c -c²)/3
= 12 + 4(16 -16 + 8c -c²)/3
= 12 + 64/3 -(c-4)²)/3
= 100/3 - (c-4)²)/3
c = 4时，S最大， 为100/3

(3)能
在（1）的两种情形下分别讨论
(a) y = (8x-x²)/3
要使其为矩形，则OB=AD，即y = 3
(8x-x²)/3 = 3
x² - 8x + 9 = 0
x = 4±√7
C(4±√7, 0) (二解均在OA上)

(b) y = 3(8-x)/x
要使其为矩形，则OB=AD，即y = 3
3(8-x)/x = 3
x = 8 - x
x = 4
C(4, 0)

Answer 2

在平面直角坐标系中，A(8,0) B（0,3），点C（x，0）是线段OA1.XY+3X=24 2.最大面积S=8*8=64（根据三角形知识，可推算出AOBD是正