题目内容：

在直角坐标系XOY中,A（0,2）B(负3,4）若C在角AOB的平分线上,向量OC的模为2,则C点坐标为?

优质解答

|OC|=2
A=(0,2),B=(-3,4)
let C be (m,n),x = OC,OB夹角
OC.OA = |OC||OA|cosx = (0,2).(m,n)
4cosx = 2n (1)
OC.OB = |OC||OB|cosx = (-3,4).(m,n)
10cosx = -3m+4n (2)
|OC| =2 => m^2+n^2 =4 (3)
(2)/(1)
5/2 = (-3m+4n)/2n
10n = -6m+8n
n = -3m (4)
Sub (4) into (3)
m^2 + 9m^2 = 4
m = √10/5(rejected) or -√10/5
n = -3m = 3√10/5
C = (-√10/5,3√10/5)