2.如图,点A的坐标为(-1,0),点B是直线y=x的一个动点,当线段AB最短时,求点B的坐标
\u5982\u56fe,\u70b9A\u7684\u5750\u6807\u4e3a(1,0),\u70b9B\u5728\u76f4\u7ebfy=-x\u4e0a\u8fd0\u52a8,\u5f53\u7ebf\u6bb5AB\u6700\u77ed\u65f6,\u70b9B\u7684\u5750\u6807\u4e3a\uff1f\u4f5cAC\u22a5BO\u4e8eC,
\u8bbe\u76f4\u7ebfAC: y=x+b
\u5c06A(1,0)\u4ee3\u5165\uff0c\u30001\uff0bb=0
b=-1
y=x-1
y=-x
\u89e3\u5f97
x=1/2 y= -1/2
\u2234B(1/2 , -1/2)
\u89e3\uff1a\u5f53AB\u4e0e\u76f4\u7ebfy=x\u5782\u76f4\u65f6\uff0cAB\u6700\u77ed\uff0e\u2235\u76f4\u7ebfy=x\u662f\u7b2c\u4e00\u3001\u4e09\u8c61\u9650\u7684\u89d2\u5e73\u5206\u7ebf\uff0c\u2234\u2220BOA=45\u00b0\uff0c\u2234\u25b3OAB\u662f\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62\uff0e\u4f5cBD\u22a5x\u8f74\uff0c\u2234DO=BD=12OA=1\uff0c\u2234B\u7684\u5750\u6807\u662f\uff081\uff0c1\uff09\uff0e\u6545\u7b54\u6848\u662f\uff1a\uff081\uff0c1\uff09\uff0e
我记得原来有个公式，现在不大记得了，用几何方法吧当AB垂直于直线y=x时，AB最短；
将ABO看成△ABO；
又∵直线方程是y=x;
故∠AOB=45°,又∠ABO=90°
所以△ABO是等腰直角三角形；
AO²=AB²+BO²=2BO²=1
BO=√2/2
B坐标为（X0，X0）;(B在直线方程y=x上，故横纵坐标相等)
X0²+Y0²=BO²=2X0²=1/2
又B在第四象限
X0=-1/2=Y0
故坐标为（-1/2，-1/2）
垂直最短 过A做AB垂直 Y=X AOB为等腰直角三角形 B坐标（-1/2， -1/2）
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