若角α的终边点P（1,-2）,则tan2α的值为多少?A.-4/3 B.4/ 3若角α的终边点P（1,-2）,则tan2α的值为多少?A.-4/3 B.4/ 3 C.3/4 D.-3/4
若角α的终边经过点P(1,-2)答案：Btana=y/x=-2tan2a=2tana/(1-tan^2a)=-4/(-3)=4/3
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其他类似问题
答案B因为公式tan(A+B)=(tanA+tanB)/(1-tanAtanB)。。又因为tana=-2带入得B
扫描下载二维码（2014o牡丹江）如图，在平面直角坐标系中，直线AB与x轴、y轴分别交于点A，B，直线CD与x轴、y轴分别交于点C，D，AB与CD相交于点E，线段OA，OC的长是一元二次方程x2-18x+72=0的两根（OA＞OC），BE=5，tan∠ABO=．
（1）求点A，C的坐标；
（2）若反比例函数y=的图象经过点E，求k的值；
（3）若点P在坐标轴上，在平面内是否存在一点Q，使以点C，E，P，Q为顶点的四边形是矩形？若存在，请写出满足条件的点Q的个数，并直接写出位于x轴下方的点Q的坐标；若不存在，请说明理由．
（1）先求出一元二次方程x2-18x+72=0的两根就可以求出OA，OC的值，进而求出点A，C的坐标；
（2）先由勾股定理求出AB的值，得出AE的值，如图1，作EM⊥x轴于点M，由相似三角形的性质就可以求出EM的值，AM的值，就可以求出E的坐标，由待定系数法就可以求出结论；
（3）如图2，分别过C、E作CE的垂线交坐标轴三个点P1、P3、P4，可作出三个Q点，过E点作x轴的垂线与x轴交与p2，即可作出Q2，以CE为直径作圆交于y轴两个点P5、P6，使PC⊥PE，即可作出Q5、Q6．
解：（1）∵x2-18x+72=0
∴x1=6，x2=12．
∵OA＞OC，
∴OA=12，OC=6．
∴A（12，0），C（-6，0）；
（2）∵tan∠ABO=，
在Rt△AOB中，由勾股定理，得
如图1，作EM⊥x轴于点M，
∴EM∥OB．
∴△AEM∽△ABO，
∴EM=12，AM=9，
∴OM=12-9=3．
∴E（3，12）．
（3）满足条件的点Q的个数是6，如图2所示，
x轴的下方的Q4（10，-12），Q6（-3，6-3）；
如图①∵E（3，12），C（-6，0），
∴CG=9，EG=12，
∴EG2=CGoGP，
∵△CPE与△PCQ中心对称，
∴CH=GP=16，QH=FG=12，
∴Q（10，-12），
如图②∵E（3，12），C（-6，0），
∴CG=9，EG=12，
∵MN=CG=，
可以求得PH=3-6，
∴Q（-3，6-3），& 一次函数综合题知识点 & “（2013o河北）如图，A（0，1），M...”习题详情
105位同学学习过此题，做题成功率81.9%
（2013o河北）如图，A（0，1），M（3，2），N（4，4）．动点P从点A出发，沿y轴以每秒1个单位长的速度向上移动，且过点P的直线l：y=-x+b也随之移动，设移动时间为t秒．（1）当t=3时，求l的解析式；（2）若点M，N位于l的异侧，确定t的取值范围；（3）直接写出t为何值时，点M关于l的对称点落在坐标轴上．
本题难度：一般
题型：解答题&|&来源：2013-河北
分析与解答
习题“（2013o河北）如图，A（0，1），M（3，2），N（4，4）．动点P从点A出发，沿y轴以每秒1个单位长的速度向上移动，且过点P的直线l：y=-x+b也随之移动，设移动时间为t秒．（1）当t=3时，求l的解析...”的分析与解答如下所示：
（1）利用一次函数图象上点的坐标特征，求出一次函数的解析式；（2）分别求出直线l经过点M、点N时的t值，即可得到t的取值范围；（3）找出点M关于直线l在坐标轴上的对称点E、F，如解答图所示．求出点E、F的坐标，然后分别求出ME、MF中点坐标，最后分别求出时间t的值．
解：（1）直线y=-x+b交y轴于点P（0，b），由题意，得b＞0，t≥0，b=1+t．当t=3时，b=4，故y=-x+4．（2）当直线y=-x+b过点M（3，2）时，2=-3+b，解得：b=5，5=1+t，解得t=4．当直线y=-x+b过点N（4，4）时，4=-4+b，解得：b=8，8=1+t，解得t=7．故若点M，N位于l的异侧，t的取值范围是：4＜t＜7．（3）如右图，过点M作MF⊥直线l，交y轴于点F，交x轴于点E，则点E、F为点M在坐标轴上的对称点．过点M作MD⊥x轴于点D，则OD=3，MD=2．已知∠MED=∠OEF=45°，则△MDE与△OEF均为等腰直角三角形，∴DE=MD=2，OE=OF=1，∴E（1，0），F（0，-1）．∵M（3，2），F（0，-1），∴线段MF中点坐标为（32，12）．直线y=-x+b过点（32，12），则12=-32+b，解得：b=2，2=1+t，解得t=1．∵M（3，2），E（1，0），∴线段ME中点坐标为（2，1）．直线y=-x+b过点（2，1），则1=-2+b，解得：b=3，3=1+t，解得t=2．故点M关于l的对称点，当t=1时，落在y轴上，当t=2时，落在x轴上．
本题是动线型问题，考查了坐标平面内一次函数的图象与性质．难点在于第（3）问，首先注意在x轴、y轴上均有点M的对称点，不要漏解；其次注意点E、F坐标以及线段中点坐标的求法．
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（2013o河北）如图，A（0，1），M（3，2），N（4，4）．动点P从点A出发，沿y轴以每秒1个单位长的速度向上移动，且过点P的直线l：y=-x+b也随之移动，设移动时间为t秒．（1）当t=3时，...
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经过分析，习题“（2013o河北）如图，A（0，1），M（3，2），N（4，4）．动点P从点A出发，沿y轴以每秒1个单位长的速度向上移动，且过点P的直线l：y=-x+b也随之移动，设移动时间为t秒．（1）当t=3时，求l的解析...”主要考察你对“一次函数综合题”
等考点的理解。
因为篇幅有限，只列出部分考点，详细请访问。
一次函数综合题
（1）一次函数与几何图形的面积问题首先要根据题意画出草图，结合图形分析其中的几何图形，再求出面积．（2）一次函数的优化问题通常一次函数的最值问题首先由不等式找到x的取值范围，进而利用一次函数的增减性在前面范围内的前提下求出最值．（3）用函数图象解决实际问题从已知函数图象中获取信息，求出函数值、函数表达式，并解答相应的问题．
与“（2013o河北）如图，A（0，1），M（3，2），N（4，4）．动点P从点A出发，沿y轴以每秒1个单位长的速度向上移动，且过点P的直线l：y=-x+b也随之移动，设移动时间为t秒．（1）当t=3时，求l的解析...”相似的题目：
[2012o湘潭o中考]已知一次函数y=kx+b（k≠0）图象过点（0，2），且与两坐标轴围成的三角形面积为2，则此一次函数的解析式为（　　）y=x+2y=x-2y=-x-2或y=x-2y=x+2或y=-x+2
[2011o牡丹江o中考]在平面直角坐标系中，点0为原点，直线y=kx+b交x轴于点A（-2，0），交y轴于点B．若△AOB的面积为8，则k的值为（　　）12-2或44或-4
[2009o鄂州o中考]如图，直线AB：y=12x+1分别与x轴、y轴交于点A，点B，直线CD：y=x+b分别与x轴，y轴交于点C，点D．直线AB与CD相交于点P，已知S△ABD=4，则点P的坐标是（　　）（3，52）（8，5）（4，3）（12，54）
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该知识点好题
1如图，直线y=34x+3交x轴于A点，将一块等腰直角三角形纸板的直角顶点置于原点O，另两个顶点M、N恰落在直线y=34x+3上，若N点在第二象限内，则tan∠AON的值为（　　）
2（2012o聊城）如图，在直角坐标系中，以原点O为圆心的同心圆的半径由内向外依次为1，2，3，4，…，同心圆与直线y=x和y=-x分别交于A1，A2，A3，A4…，则点A30的坐标是（　　）
3（2011o仙桃）如图，已知直线l：y=√33x，过点A（0，1）作y轴的垂线交直线l于点B，过点B作直线l的垂线交y轴于点A1；过点A1作y轴的垂线交直线l于点B1，过点B1作直线l的垂线交y轴于点A2；…；按此作法继续下去，则点A4的坐标为（　　）
该知识点易错题
1已知直线y=-√3x+√3与x轴，y轴分别交于A，B两点，在坐标轴上取一点P，使得△PAB是等腰三角形，则符合条件的点P有（　　）个．
2一次函数y=-2x+4与x，y轴分别交于A，B点，且C是OA的中点，则在y轴上存在（　　）个点D，使得以O，D，C为顶点的三角形与以O，A，B为顶点的三角形相似．
3一次函数y=54x-15的图象与x轴、y轴分别交于点A、B，O为坐标原点，则在△OAB内部（包括边界），纵坐标、横坐标都是整数的点共有（　　）
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