如图,在直角坐标系中,点A的坐标为(-2,0),连接OA,将线段OA绕原点O顺时针旋转120°,得到线段OB.

2025-06-22 20:38:21
推荐回答(1个)
回答1:

1)
B(1, √3)
y=a(x+2)x
 
√3=a*3
a=√3/3
y=√3/3x^2+2√3/3x
 
(2)
A(-2,0),B(1, √3)
AB: y=√3/3x+2√3/3
 
x= - 1 ,y= √3/3
P(- 1, √3/3)
 
(3)
y=√3/3x^2+2√3/3x
x= - 1 ,y= - √3/3
M(- 1, - √3/3)
HM=HP,HA=HO
 
△HMO≌△HPA
(1)
B(1, √3)
y=a(x+2)x
 
√3=a*3
a=√3/3
y=√3/3x^2+2√3/3x
 
(2)
A(-2,0),B(1, √3)
AB: y=√3/3x+2√3/3
 
x= - 1 ,y= √3/3
P(- 1, √3/3)
 
(3)
y=√3/3x^2+2√3/3x
x= - 1 ,y= - √3/3
M(- 1, - √3/3)
HM=HP,HA=HO
 
△HMO≌△HPA
所以OM//=PA
M(- 1, - √3/3), A、P、O、M为顶点的四边形是平行四边形
所以OM//=PA
M(- 1, - √3/3), A、P、O、M为顶点的四边形是平行四边形