Question

如图,直线y=-根号3/3x+1与x轴、y轴分别交于点A、B,以线段AB为直角边在第一象限作等腰直角△ABC,∠BAC=

Answer 1

y = -(√3)x/3 + 1

x = 0, y = 1, B(0, 1)

y = 0, x = √3, A(√3, 0)

AB = √(3 + 1) = 2

AC = AB = 2 即C与直线的距离为2,

△ABP的面积与△ABC的面积相等, 二者有共同的底(AB),所以只须AB上的高相等即可.即P与直线的距离也是2.

y = -(√3)x/3 + 1可变为(√3)x/3 + y - 1 = 0

P与直线的距离d = 2 = |(√3)a/3 + 1/2 -1|/√(1/3 + 1) = |a/√3 -1/2|/√(4/3)

|a/√3 -1/2| = 4/√3

|a - √3/2| = 4

a < 0, -a+√3/2 = 4

a = √3/2 -4