Question

设直线y=x-2与抛物线y=-x平方相交于A,B两点,求交点A,B的坐标和线段AB的长

Answer 1

y = x - 2

y = -x^2

x - 2 = -x^2

x^2 + x - 2 = 0

x = 1或-2

代入,得y = -1或-4

所以A、B坐标分别为(1,-1),(-2,-4)

AB = 根号((1-(-2))^2 + (-1-(-4))^2)

= 根号(9+9)

= 3×根号2