在平面直角坐標(biāo)系中,A(-4,-2),B(-2,-2),C(-1,0)
(1)將△ABC繞C點順時針旋轉(zhuǎn)90°,得△A1B1C,則點A1的坐標(biāo)為______.
(2)將△A1B1C向右平移6個單位得△A2B2C2,則點B2的坐標(biāo)為______.
(3)從△ABC到△A2B2C2能否看作是繞某一點作旋轉(zhuǎn)變換?若能,則旋轉(zhuǎn)中心坐標(biāo)為______在旋轉(zhuǎn)變換中AB所掃過的面積為______.
【答案】
分析:(1)在(-1,-2)處取點D,可知A,B,D三點位于同一直線上,且△ACD為直角三角形,即∠ADC=90°.我們讓△ACD繞C點旋轉(zhuǎn),易知CD與x軸重合,A
1D∥y軸,即A′橫坐標(biāo)的數(shù)值等于CD的長度加上OC的長度,縱坐標(biāo)等于AD的長度,又A
1位于第二象限,故A
1的坐標(biāo)為(-3,3).
(2)由(1)可知,B
1的坐標(biāo)為(-3,1),A
1B
1C向右平移6個單位得△B
2C
2,B
1的橫坐標(biāo)向右平移6個單位,即B
2的橫坐標(biāo)為-3+6=3,即點B
2的坐標(biāo)為(3,1).
(3)要求其中心,我們可以連接AA
2,CC
2,分別求他們的中垂線的方程,他們的交點就是旋轉(zhuǎn)中心,易知CC
2的中垂線為x=2,AA
2的斜率為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/0.png)
,其中點Q坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/1.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/2.png)
),所以其中垂線的方程為5y+7x+1=0,與x=2聯(lián)立,解得交點P坐標(biāo)為(2,-3).
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/3.png)
的面積等于扇形PAB的面積減去△PAB的面積,易知PA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/4.png)
,PQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/5.png)
,可知∠APQ=60°,即∠APA
2=120°.所以
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/6.png)
=S
扇PAA2-S
△APQ.同理可求出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/7.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/8.png)
.即S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/9.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/10.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/11.png)
.
解答:解:(1)取點D(-1,-2),可知A,B,D三點同一直線上,所以△ACD為直角三角形(∠ADC=90°),△ACD繞C點旋轉(zhuǎn),易知CD與x軸重合,A
1D∥y軸,即A′橫坐標(biāo)的數(shù)值等于CD的長度加上OC的長度,縱坐標(biāo)等于AD的長度,又A
1位于第二象限,故A
1的坐標(biāo)為(-3,3).A
1(-3,3);
(2)由(1)可知,B
1的坐標(biāo)為(-3,1),A
1B
1C向右平移6個單位得△B
2C
2,B
1的橫坐標(biāo)向右平移6個單位,即B
2的橫坐標(biāo)為-3+6=3,即點B
2的坐標(biāo)為(3,1).B
2(3,1);
(3)連接AA
2,CC
2,易知AA
2的斜率為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/12.png)
,其中點Q的坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/13.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/14.png)
),所以其中垂線的方程為5y+7x+1=0,CC
2的中垂線為x=2,與x=2聯(lián)立,解得交點P坐標(biāo)為(2,-3).易知PA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/15.png)
,PQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/16.png)
,可知∠APQ=60°,即∠APA
2=120°.所以
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/17.png)
=S
扇PAA2-S
△APQ.同理可求出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/19.png)
.即S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/20.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/21.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021230147290474221/SYS201310212301472904742020_DA/22.png)
,經(jīng)計算S=5π.
點評:此題較為復(fù)雜,是對學(xué)生對旋轉(zhuǎn)問題的靈活運用以及對學(xué)生要求一定的計算能力.