世纪金榜高中全程复习方略详细答案8.2.ppt

第二节 直线的交点坐标与距离公式 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 三年2考 高考指数: 1.能用解方程组的方法求两条相交直线的交点坐标； 2.掌握两点间的距离公式、点到直线的距离公式，会求两条平 行直线间的距离. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1.两点间距离公式、点到直线的距离公式,两平行线间的距离 公式是高考的重点； 2.常与圆、椭圆、双曲线、抛物线交汇命题； 3.多以选择题和填空题为主，有时与其他知识点交汇，在解答 题中考查. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1.两条直线的交点 直线l1：A1x+B1y+C1=0与l2：A2x+B2y+C2=0的公共点的坐标与方 程组 的解一一对应. 相交方程组有_，交点坐标就是方程组的解； 平行方程组_；重合方程组有_. 唯一解 无解无数组解 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【即时应用】 (1)思考：如何用两直线的交点判断两直线的位置关系？ 提示：当两直线有一个交点时，两直线相交；没有交点时，两 直线平行；有无数个交点时，两直线重合. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (2)直线l1：5x+2y-6=0与l2：3x-5y-16=0的交点P的坐标是_. 【解析】由直线l1与l2所组成的方程组 得： ，直线l1：5x+2y-6=0与l2：3x-5y-16=0的交点P的 坐标是(2,-2). 答案：(2,-2) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (3)直线l1：5x+2y-6=0与l2：5x+2y-16=0的位置关系是_. 【解析】由直线l1与l2所组成的方程组 无解，直线l1与l2平行. 答案：平行 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.距离 两条平行线 Ax+By+C1=0与Ax +By+C2 =0间的距离 点P0(x0,y0)到直线 l:Ax +By +C =0的距 离 点P1(x1,y1), P2(x2,y2)之间的距 离 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【即时应用】 (1)原点到直线x+2y-5=0的距离是_； (2)已知A(a,-5)，B(0,10)，|AB|=17，则a=_； (3)两平行线y=2x与2x-y=-5间的距离为_. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解析】(1)因为 (2)依题设及两点间的距离公式得： ,解得：a=8； (3)因为两平行线方程可化为：2x-y=0与2x-y+5=0. 因此，两平行线间的距离为： 答案：(1) (2)8 (3) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 两直线的交点问题 【方法点睛】 1.两直线交点的求法 求两直线的交点坐标，就是解由两直线方程组成的方程组，以 方程组的解为坐标的点即为交点. 2.过直线A1x+B1y+C1=0与A2x+B2y+C2=0交点的直线系方程 A1x+B1y+C1+(A2x+B2y+C2)=0.(不包括直线A2x+B2y+C2=0) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例1】(1)求经过直线x+y+1=0与直线x-y+3=0的交点，且也经 过点A(8,-4)的直线方程为_； (2)已知两直线l1：mx+8y+n=0与l2：2x+my-1=0，若l1与l2相交， 求实数m、n满足的条件. 【解题指南】(1)可求出两直线的交点坐标，用两点式解决； 也可用过两直线交点的直线系解决；(2)两直线相交可考虑直 线斜率之间的关系，从而得到m、n满足的条件. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【规范解答】(1)方法一：因为直线x+y+1=0与直线x-y+3=0的 交点坐标为(-2,1)，直线又过A(8,-4)，所以所求直线方程 为： ，即x+2y=0； 方法二：设过直线x+y+1=0与直线x-y+3=0的交点的直线方程为 x+y+1+(x-y+3)=0， 又因为直线过A(8,-4)，所以8-4+1+(8+4+3)=0， 解得: ,所以，所求直线方程为x+2y=0. 答案：x+2y=0 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (2)因为两直线l1：mx+8y+n=0与l2：2x+my-1=0相交，因此，当 m=0时，l1的方程为 ，l2的方程为 ，两直线相交，此 时，实数m、n满足的条件为m=0，nR；当m0时，两直线 相交， ，解得m4，此时，实数m、n满足的条件为 m4，nR. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【互动探究】本例(1)中的“且也经过点A(8,-4)”改为“与直 线2x-y=0垂直”,求该直线方程. 【解析】方法一：因为直线x+y+1=0与直线x-y+3=0的交点坐标 为(-2,1)，又直线与直线2x-y=0垂直，所以所求直线的斜率 ，因此所求直线方程为：y-1= (x+2)，即x+2y=0. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 方法二：设过直线x+y+1=0与直线x-y+3=0的交点的直线方程为 x+y+1+(x-y+3)=0，即 (1+)x+(1-)y+1+3=0, 又因为直线与直线2x-y=0垂直，所以所求直线的斜率 即有 解得: ,所以，所求直线方程为x+2y=0. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【反思感悟】1.本例(1)中是求直线方程,其关键是寻找确定 直线的两个条件,可以直接求交点,利用两点式得出方程,此法 要注意两点的纵(或横)坐标相同时,两点式方程不适用,也可以 利用直线系方程求解,其关键是利用已知点求的值; 2.考查两直线相交的条件,即斜率不等或有一条直线的斜率不 存在. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【变式备选】当m为何值时，三条直线l1：4x+y-3=0与l2： x+y=0,l3：2x-3my-4=0能围成一个三角形? 【解析】三条直线能围成三角形即三条直线两两相交且不共 点，所以 解得： Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 又因为l1：4x+y-3=0与l2：x+y=0的交点为(1,-1)，所以2+3m- 40，解得m ； 当m=0时，l3:2x-4=0,l1:4x+y-3=0,l2:x+y=0,l1与l3的交点为 (2,-5)， l1与l2的交点为(1,-1),l2与l3的交点为(2,-2)， 能构成三角形，符合题意. 综上可知： Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 距离公式的应用 【方法点睛】 1.两点间的距离的求法 设点A(xA,yA),B(xB,yB)， 特例：ABx轴时，|AB|=|yA-yB| ABy轴时，|AB|=|xA-xB|. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.点到直线的距离的求法 可直接利用点到直线的距离公式来求，但要注意此时直线方程 必须为一般式. 3.两平行直线间的距离的求法 (1)利用“化归”法将两条平行线间的距离转化为一条直线上 任意一点到另一条直线的距离. (2)利用两平行线间的距离公式. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【提醒】应用两平行线间的距离公式求距离时,要注意两平行 直线方程中x、y的系数必须相等. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例2】已知三条直线l1:2x-y+a=0(a0),l2:-4x+2y+1=0和 l3:x+y-1=0，且l1与l2的距离是 (1)求a的值； (2)能否找到一点P,使P同时满足下列三个条件： P是第一象限的点； P点到l1的距离是P点到l2的距离的 ; P点到l1的距离与P点到l3的距离之比是 .若能，求P点坐 标；若不能，说明理由. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解题指南】(1)由l1与l2的距离及两平行线之间的距离公式， 可得关于a的方程，解方程即可得出a的值； (2)由点P(x0,y0)满足条件可得出关于x0、y0的方程组，解 方程组，即可求出点P的坐标，注意验证是否适合条件. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【规范解答】(1)l2为2x-y- =0, l1与l2的距离为 a0,a=3. (2)设存在第一象限的点P(x0,y0)满足条件,则P点在与l1、l2平 行的直线l:2x-y+c=0上且 ,即 或 2x0-y0+ =0或2x0-y0+ =0. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 若P点满足条件，由点到直线的距离公式有： 即|2x0-y0+3=|x0+y0-1|, x0-2y0+4=0或3x0+2=0. P在第一象限，3x0+2=0不可能. 联立方程 和x0-2y0+4=0, 解得 (舍去)， Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 由 存在P( )同时满足条件. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【反思感悟】在解答本题时，首先要根据题设条件，由点到 直线的距离公式、两平行线间的距离公式得出方程(组)，这是 很关键的问题；另外，还要注意每种距离公式所要求的条件， 以防漏解、错解. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【变式训练】已知A(4,-3),B(2,-1)和直线l:4x+3y-2=0，在坐 标平面内求一点P，使PA=|PB|，且点P到直线l的距离为2. 【解析】设点P的坐标为(a,b). A(4，-3)，B(2,-1)， 线段AB的中点M的坐标为(3，-2)， 线段AB的垂直平分线方程为y+2=x-3, 即x-y-5=0. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 由题意知点P(a,b)在上述直线上，a-b-5=0. 又点P(a,b)到直线l：4x+3y-2=0的距离为2， ,即4a+3b-2=10, 联立可得 所求点P的坐标为(1，-4)或( ). Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【变式备选】过点P(-1,2)引一直线，两点A(2,3)，B(-4,5)到 该直线的距离相等，求这条直线的方程. 【解析】方法一：当斜率不存在时，过点P(-1,2)的直线方程 为：x=-1，A(2,3)到x=-1的距离等于3，且B(-4,5)到x=-1的距 离也等于3，符合题意； 当直线的斜率存在时，设斜率为k，过点P(-1,2)的直线方程 为：y-2=k(x+1)，即kx-y+k+2=0， Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 依题设知： 解上式得： 所以，所求直线方程为：x+3y-5=0； 综上可知,所求直线方程为x=-1或x+3y-5=0. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 方法二：依题设知：符合题意的直线共有两条，一条是过点 P(-1,2)与AB平行的直线，另一条是过点P及AB中点的直线. 因为A(2,3)，B(-4,5)，所以 因此，过点P与AB平行的直线的方程为： y-2= (x+1),即x+3y-5=0; 又因为A(2,3)，B(-4,5)的中点坐标D(-1,4)， 所以过点P及AB中点的直线方程为x=-1； 综上可知,所求直线方程为x=-1或x+3y-5=0. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 对称问题 【方法点睛】 1.对称中心的求法 若两点A(x1,y1)、B(x2,y2)关于点P(a,b)对称，则由中点坐标 公式求得a、b的值，即 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.轴对称的两个公式 若两点M(x1,y1)、N(x2,y2)关于直线l：Ax+By+C=0(A0)对称， 则线段MN的中点在对称轴l上，而且连接MN的直线垂直于对称 轴l.故有 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3.对称问题的类型 (1)点关于点对称；(2)点关于直线对称； (3)直线关于点对称；(4)直线关于直线对称. 以上各种对称问题最终化归为点关于点对称、点关于直线对称 . 4.对称问题的具体应用 (1)在直线上求一点，使它到两定点距离之和最小问题 当两定点分别在直线的异侧时，两点连线与直线的交点即为 所求； Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 当两定点在直线的同一侧时，可借助于点关于直线对称，将 问题转化为情形来解决. (2)在直线上求一点，使它到两定点距离之差的绝对值最大问 题 当两定点在直线的同一侧时，利用三角形的两边之差小于第 三边，可知两定点的连线与直线的交点即为所求; 当两定点分别在直线的异侧时，可借助于点关于直线对称， 将问题转化为情形解决. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例3】求直线a：2x+y-4=0关于直线l：3x+4y-1=0对称的直线 b的方程. 【解题指南】本题实质上是求直线的方程，可设法找到两个点 的坐标，再由两点式即可求出方程；本题还可利用求曲线方程 的方法求解，设所求曲线上任意一点，由该点关于直线l的对称 点在已知曲线上，即可求得. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【规范解答】方法一：由 解得直线a与l的交点 E(3,-2)，E点也在直线b上. 在直线a：2x+y-4=0上取一点A(2,0)，设A点关于直线l的对称点 B的坐标为(x0,y0)， 由 ,解得B( ). 由两点式得直线b的方程为 即2x+11y+16=0. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 方法二：设直线b上的动点P(x,y)关于l：3x+4y-1=0的对 称点为Q(x0,y0).则 解上式得： 由于Q(x0,y0)在直线a：2x+y-4=0上，则 化简得2x+11y+16=0是所求的直线b的方程. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【反思感悟】1.此题是求直线关于直线对称的直线方程问题 ，通过求解本题，我们可体会到求直(曲)线的对称直(曲)线方 程时可以转化为求点的对称点坐标来求解. 2.利用两点式求直线方程要注意两点横坐标相等或纵坐标相等 的情形，此时可直接写出直线方程. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【变式训练】(1)在直线l：3x-y-1=0上求一点P，使得P到 A(4,1)和B(0,4)的距离之差最大； (2)在直线l：3x-y-1=0上求一点Q，使得Q到A(4，1)和C(3，4) 的距离之和最小. 【解析】(1)如图甲所示，设点B关于l 的对称点为B，连接AB并延长交l于 P，此时的P满足|PA|-|PB|的值最大. 设B的坐标为(a,b), 则kBBkl=-1. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 即 a+3b-12=0, 又由于线段BB的中点坐标为( ),且在直线l上， ,即3a-b-6=0. 联立，解得a=3,b=3,B(3,3). 于是AB的方程为 即2x+y-9=0,解 所求P点的坐标为(2，5). Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (2)如图乙所示，设C关于l的对称点为C，连接AC与l交 于点Q，此时的Q满足|QA|+|QC|的值最小. 设C的坐标为(x,y), 解得 C( ). Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 由两点式得直线AC的方程为 即19x+17y-93=0. 解 所求Q点的坐标为 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【创新探究】新定义下的直线方程问题 【典例】(2012上海模拟)在平面直角坐标系中，设点 P(x,y)，定义OP=|x|+|y|，其中O为坐标原点. 对于以下结论：符合OP=1的点P的轨迹围成的图形的面积 为2； 设P为直线 x+2y-2=0上任意一点，则OP的最小值为1； 其中正确的结论有_(填上你认为正确的所有结论的序号) . Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【解题指南】根据新定义，讨论x的取值，得到y与x的分段 函数关系式，画出分段函数的图象，即可求出该图形的面积； 认真观察直线方程，可举一个反例，得到OP的最小值为1 是假命题. 【规范解答】由OP=1，根据新定义 得：|x|+|y|=1,上式可化为：y=-x+1 (0x1)，y=-x-1(-1x0)，y=x+ 1(-1x0)，y=x-1(0x1),画出 图象如图所示： Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 根据图形得到：四边形ABCD为 边长是 的正方形，所以面积 等于2，故正确； 当点P为( ，0)时，OP =|x|+|y|= +01, 所以OP的最小值不为1，故错误； 所以正确的结论有：. 答案： x-11 y o -1 1 A B C D Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【阅卷人点拨】通过对本题的深入研究，我们可以得到以下创 新点拨和备考建议： 创 新 点 拨 本题有以下两处创新点 (1)考查内容的创新，对解析几何问题与函数知识巧 妙结合进行考查. (2)考查对新定义、新概念的理解与运用的同时考查 思维的创新,本题考查了学生的发散思维，思维方向 与习惯思维有所不同. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 备 考 建 议 解决新概念、新定义的创新问题时，要注意以下几点 ： (1)充分理解概念、定理的内涵与外延； (2)对于新概念、新结论要具体化，举几个具体的例 子，代入几个特殊值； (3)注意新概念、新结论正用怎样，逆用又将如何， 变形将会如何. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1.(2012福州模拟)直线l1的斜率为2,l1l2，直线l2过点(-1,1) 且与y轴交于点P，则P点坐标为( ) (A)(3，0) (B)(-3，0) (C)(0，-3) (D)(0，3) 【解析】选D.点P在y轴上，设P(0,y), 又kl1=2,l1l2,kl2= y=3,P(0,3). Evaluation only