圆x^2+y^2=25,△ABC内接此圆,点A的坐标为(3,4),直线AB与直线AC的倾斜角互补,求证：直线BC的斜率为定值.

Question

圆x^2+y^2=25,△ABC内接此圆,点A的坐标为(3,4),直线AB与直线AC的倾斜角互补,求证：直线BC的斜率为定值.

Answer 1

直线AB与直线AC的倾斜角互补,可得：AB=AC
设点B的坐标为(a,b),点C的坐标为(s,t)则有：
直线BC的斜率为：k=(t-b)/(s-a)
a^2+b^2=25`````````````````````````````````````````1
s^2+t^2=25``````````````````````````````````````````2
(3-a)^2+(4-b)^2=(3-s)^2+(4-t)^2`````````````````3
联立1、2、3,化简整理得：
3a+4b=3s+4t
即：3(a-s)=4(t-b)
可得：k=(t-b)/(s-a)=-3/4 为定值,得证：