二次函数y＝-1/2x²＋3/2+m-2的图像与x轴交于A,B两点与y轴交于C,且∠ACB＝90°

1.
设A(a,0),B(b,0),a < 0,b > 0
y = (-1/2)(x - a)(x - b) = -x²/2 + (a + b)x/2 -ab/2
-ab/2 = m - 2,ab = 2(2 - m) (i)
AC的斜率p = (m - 2 - 0)/(0 - a) = (2 - m)/a
BC的斜率q = (m - 2 - 0)/(0 - b) = (2 - m)/b
∠ACB＝90°,pq = -1 = (m - 2)²/(ab)
(m - 2)² = -ab =2(m - 2)
(m - 2)(m - 2 - 2) = 0
m = 4 (舍去m = 2,此时C为原点)
y = -x²/2＋3x/2+ 2
2.
y = (-1/2)(x + 1)(x - 4)
A(-1,0),B(4,0)
OB = 4,OC = 2
三角形BOC面积S = (1/2)OB*OC = (1/2)*4*2 = 4
截得的三角形面积 = S/4 = 1
(1)
作一条与y轴平行的直线,与x轴交于M(m,0),m > 0:与BC交于N(m,n)
显然三角形BMN与三角形BCA相似
BC解析式:x/4 + y/2 = 1
x = m,y = (4 - m)/2
N(m,(4 - m)/2)
三角形BMN面积 = (1/2)*MB*MN = (1/2)(4 - m)(4 - m)/2 = (m - 4)²/4 = 1
m - 4 = ± 2
m = 2 (舍去m = 6 > 4)
三个顶点M(2,0),N(2,1),B(4,0)
(2)
作一条与x轴平行的直线y = p,0 < p < 2,直线与AC交于M(m,p),m < 0:与BC交于N(n,p)
BC解析式:x/4 + y/2 = 1,y = p,x = 2(2 - p),N(2(2 - p),p)
AC解析式:x/(-1) + y/2 = 1,y = p,x = (p - 2)/2,M(p - 2)/2,p)
显然三角形CMN与三角形CAB相似
MN = 2(2 - p) - (p - 2)/2 = 5(2 - p)/2
MN上的高h = C的纵坐标 - M的纵坐标 = 2 - p
三角形CMN面积 = (1/2)MN*h = 5(2 -p)²/4 = 1
p - 2 = ± 2√5/5
p = 2 - 2√5/5 (舍去p = 2 + 2√5/5 > 2)
M(-√5/5,2 - 2√5/5),N(4√5/5,2 - 2√5/5),C(0,2)

1.根据RT三角形斜边中线等于斜边一半。c点坐标（0，m-2），斜边中点（3/2,0）根据对称轴得出。-1/2x²＋3/2+m-2=0的两根之和3，两根之积-2（m-2)
得出4（（m-2）²+9/4）=9-4（-2（m-2) ）得4（m-2）²=8（m-2) 得m=2（不合题意）或m=4。

2.代入m=4，得出A（-1,0）...