（2013?房山区二模）如图，直线AB过点A，且与y轴交于点B．（1）求直线AB的解析式；（2）若P是直线AB上一

\uff082012?\u623f\u5c71\u533a\u4e8c\u6a21\uff09\u5982\u56fe\uff0c\u5e73\u9762\u76f4\u89d2\u5750\u6807\u7cfb\u4e2d\uff0c\u76f4\u7ebfAB\u4e0ex\u8f74\u4ea4\u4e8e\u70b9A\uff082\uff0c0\uff09\uff0c\u4e0ey\u8f74\u4ea4\u4e8e\u70b9B\uff0c\u70b9D\u5728\u76f4\u7ebfAB\u4e0a

\uff081\uff09\u8bbe\u76f4\u7ebfAB\u7684\u89e3\u6790\u5f0f\u4e3a\uff1ay=kx+b\uff0c\u2235\u70b9A\uff082\uff0c0\uff09\uff0c\u70b9D\uff081\uff0c3\uff09\uff0c\u22342k+b\uff1d0k+b\uff1d3\uff0c\u89e3\u5f97\uff1ak\uff1d?3b\uff1d23\uff0c\u2234\u76f4\u7ebfAB\u7684\u89e3\u6790\u5f0f\u4e3a\uff1ay=-3x+23\uff1b\uff082\uff09\u2235\u76f4\u7ebfAB\u7684\u89e3\u6790\u5f0f\u4e3a\uff1ay=-3x+23\uff1b\u2234\u70b9B\u7684\u5750\u6807\u4e3a\uff080\uff0c23\uff09\uff0c\u2234OA=2\uff0cOB=23\uff0c\u2234\u5728Rt\u25b3AOB\u4e2d\uff0ctan\u2220BAO=OBOA=3\uff0c\u2234\u2220BAO=60\u00b0\uff0c\u5f53\u76f4\u7ebfAB\u7ed5\u70b9A\u9006\u65f6\u9488\u65cb\u8f6c30\u00b0\u4ea4y\u8f74\u4e8e\u70b9C\uff0c\u2234\u2220CAO=\u2220BAO-30\u00b0=30\u00b0\uff0c\u5728Rt\u25b3AOC\u4e2d\uff0cOC=OA?tan30\u00b0=2\u00d733=233\uff0c\u2234\u70b9C\u7684\u5750\u6807\u4e3a\uff080\uff0c233\uff09\uff0c\u8bbe\u6240\u5f97\u76f4\u7ebf\u4e3ay=mx+233\uff0c\u2235A\uff082\uff0c0\uff09\uff0c\u22340=2m+233\uff0c\u89e3\u5f97\uff1am=-<div style="width: 6px; background-image: url(http://hiphotos.baidu.com/zhidao/pic/item/dc54564e9258d1099efce906d258ccbf6c814d62.jpg); background-attachment: initial; background-origin: initial; background-clip: initial; background-color: initial; overflow-x: hidden; overflow-y: hidden; height: 1

\u8bbe\u76f4\u7ebfAB\u7684\u89e3\u6790\u5f0f\u4e3ay=kx+b\uff08k\u22600\uff09\uff0c\u2235\u70b9A\uff08-2\uff0c0\uff09\uff0cB\uff080\uff0c1\uff09\u5728\u76f4\u7ebfAB\u4e0a\uff0c\u2234?2k+b\uff1d0b\uff1d1\uff0c\u89e3\u5f97k\uff1d12b\uff1d1\uff0c\u2234\u76f4\u7ebfAB\u7684\u89e3\u6790\u5f0f\u4e3ay=12x+b\uff0c\u2234AB=(?2?0)2+(0?1)2=5\uff0c\u8bbeB1\uff08x1\uff0c12x1+b\uff09\uff0cB2\uff08x2\uff0c12x2+b\uff09\uff0cB3\uff08x3\uff0c12x3+b\uff09\u7684\u5750\u6807\uff0c\u2235BB1=AB\uff0cB1B2=BB1\uff0cB2B3=B1B2\uff0c\u2234\uff08x1-0\uff092+\uff0812x1+1-1\uff092=\uff085\uff092\uff0c\u89e3\u5f97x=2\u6216x=-2\uff08\u820d\u53bb\uff09\uff0c\u2234B1\uff082\uff0c2\uff09\uff0c\u540c\u7406\u53ef\u5f97B2\uff084\uff0c3\uff09\uff0cB3\uff086\uff0c4\uff09\uff0c\u2234S\u77e9\u5f62OA1B1C1=2\u00d72=2\u00d7\uff081+1\uff09=4\uff1bS\u77e9\u5f62OA2B2C2=4\u00d73=2\u00d72\u00d7\uff082+1\uff09=12\uff1bS\u77e9\u5f62OA3B3C3=6\u00d74=2\u00d73\u00d7\uff083+1\uff09=14\uff0c\u2234\u7b2cn\u5404\u77e9\u5f62\u7684\u9762\u79ef=2n\uff08n+1\uff09=2n2+2n\uff0e\u6545\u7b54\u6848\u4e3a\uff1a24\uff1b2n2+2n\uff0e

解：（1）由图可知：A（-3，-3），B（0，3）
设直线AB的解析式为y=kx+b（k≠0）
则

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