Question

在直线l上是否存在一点p,使以P,A,B,D为顶点的四边形是平行四边形

Answer 1

y=x²-2x+c

y=(x-1)²+c-1

A(1,c-1)代入y=x-5

c-1=1-5

c=-3

y=x²-2x-3

y=(x-3)(x+1)

A(1,-4) B(0,-3) D(3,0)

AB斜率k1=-1

PD方程y=-x+3

y=-x+3和y=x-5的交点

-x+3=x-5

x=4

y=-1

P(4,-1)

错点如下

P为(a,5-a),则AP²=(a-1)²+(9-a)²

P是(a,a-5)

AP²是=(a-1)²+(a-1)²

AP=BD

BD=3²+3²=36

(a-1)²+(a-1)²=36

a²-2a+1=9

a²-2a+8=0

(a-4)(a+2)=0

a=4 a=-2

取a=4