Question

19£®Èçͼ1£¬Æ½ÃæÖ±½Ç×ø±êϵxOyÖУ¬Õý·½ÐÎABCDµÄ±ßABÔÚxÖáÉÏ£¬µãOÊÇABµÄÖе㣬ֱÏßl£ºy=kx-2k+4¹ý¶¨µãC£¬½»xÖáÓÚµãE£®
£¨1£©ÇóÕý·½ÐÎABCDµÄ±ß³¤£»
£¨2£©Èçͼ2£¬µ±k=-$\frac{4}{3}$ʱ£¬¹ýµãC×÷FC¡ÍCE£¬½»ADÓÚµãF£¬Á¬½ÓEF£¬BDÏཻÓÚµãH£¬BD½»yÖáÓÚG£¬ÇóÏ߶ÎGHµÄ³¤£®
£¨3£©Èçͼ3£¬ÔÚÖ±ÏßlÉÏÓÐÒ»µãN£¬CN=$\frac{1}{2}AB$£¬Á¬½ÓAN£¬µãMΪANµÄÖе㣬Á¬½ÓBM£¬ÇóÏ߶ÎBMµÄ³¤¶ÈµÄ×îСֵ£¬²¢Çó³ö´ËʱµãNµÄ×ø±ê£®

Answer 1

·ÖÎö £¨1£©ÓÉy=kx-2k+4£¬¿ÉµÃy-4=k£¨x-2£©£¬ÓÉy=kx-2k+4¹ý¶¨µã£¬ÔòxÓëyµÄÖµÓëkÎ޹أ¬¿ÉµÃ$\left\{\begin{array}{l}{x-2=0}\\{y-4=0}\end{array}\right.$£¬½âµÃ$\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$£¬½ø¶øµÃ³öCµãµÄ×ø±ê£¬¼´¿ÉµÃ³öÕý·½ÐÎABCDµÄ±ß³¤Îª4£¬
£¨2£©ÓÉk=-$\frac{4}{3}$ʱ£¬µÃ³öÖ±ÏßlµÄ½âÎöʽΪy=-$\frac{4}{3}$x+$\frac{20}{3}$£¬´Ó¶øµÃ³öµãEµÄ×ø±ê£¬ÓÉFC¡ÍCE£¬¡ÏDCB=90¡ã£¬¡ÏDCF=¡ÏBCE£¬¿ÉµÃ¡÷DCF¡Õ¡÷BCE£¨ASA£©£¬ÓÉDF=BE=5-2=3£¬AF=1£¬µÃ³öµãF£¨-2£¬1£©£¬ÓÉÖ±ÏßEFµÄ½âÎöʽΪy=-$\frac{1}{7}$x+$\frac{5}{7}$£¬Ö±ÏßBDµÄ½âÎöʽΪy=-x+2£¬ÁªÁ¢µÃµÃ³öG£¨0£¬2£©£¬ÀûÓÃÁ½µã¼äµÄ¾àÀë¿ÉµÃ³öGHµÄÖµ£¬
£¨3£©ÔÚxÖáÉϽØÈ¡BP=AB£¬Á¬½ÓNP¡¢CP£¬ÓÉCN=$\frac{1}{2}$AB=2£¬CP=4$\sqrt{2}$£¬¿ÉµÃNP¡ÜCP-CN=4$\sqrt{2}$-2£¬ËùÒÔµ±C¡¢N¡¢PÈýµã¹²Ïßʱ£¬È¡µÃ×î´óÖµ£¬ÓÖÓÉMΪANµÄÖе㣬BΪAPµÄÖе㣬µÃ³öÏ߶ÎBMµÄ³¤¶ÈµÄ×îСֵΪBM=$\frac{1}{2}$NP¡Ü2$\sqrt{2}$-1£¬ÀûÓÃÏàËÆÈý½ÇÐÎÏàËƱȿɵóöNµÄ×ø±ê£®

½â´ð ½â£º£¨1£©ÓÉy=kx-2k+4£¬µÃy-4=k£¨x-2£©£¬
¡ßÖ±Ïßl£ºy=kx-2k+4¹ý¶¨µã£¬ÔòxÓëyµÄÖµÓëkÎ޹أ¬
¡à$\left\{\begin{array}{l}{x-2=0}\\{y-4=0}\end{array}\right.$£¬½âµÃ$\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$£¬
¡àC£¨2£¬4£©£¬
¡àÕý·½ÐÎABCDµÄ±ß³¤Îª4£¬
£¨2£©µ±k=-$\frac{4}{3}$ʱ£¬Ö±ÏßlµÄ½âÎöʽΪy=-$\frac{4}{3}$x+$\frac{20}{3}$£¬
µ±y=0ʱ£¬x=5£¬
¡àE£¨5£¬0£©£¬
¡ßFC¡ÍCE£¬¡ÏDCB=90¡ã£¬
¡à¡ÏDCF=¡ÏBCE£¬
ÔÚ¡÷DCFºÍ¡÷BCEÖУ¬
$\left\{\begin{array}{l}{¡ÏDCF=¡ÏBCE}\\{CD=CB}\\{¡ÏCDF=¡ÏCBE}\end{array}\right.$£¬
¡à¡÷DCF¡Õ¡÷BCE£¨ASA£©£¬
¡àDF=BE=5-2=3£¬AF=1£¬
¡àF£¨-2£¬1£©
¡àÖ±ÏßEFµÄ½âÎöʽΪy=-$\frac{1}{7}$x+$\frac{5}{7}$£¬
¡ßB£¨2£¬0£©£¬D£¨-2£¬4£©£¬
¡àÖ±ÏßBDµÄ½âÎöʽΪy=-x+2£¬
ÁªÁ¢µÃ$\left\{\begin{array}{l}{y=-\frac{1}{7}x+\frac{5}{7}}\\{y=-x+2}\end{array}\right.$£¬½âµÃ$\left\{\begin{array}{l}{x=\frac{3}{2}}\\{y=\frac{1}{2}}\end{array}\right.$£¬
¡ßG£¨0£¬2£©£¬
¡àGH=$\sqrt{£¨\frac{3}{2}-0£©^{2}+£¨\frac{1}{2}-2£©^{2}}$=$\frac{3\sqrt{2}}{2}$£¬
£¨3£©Èçͼ3£¬ÔÚxÖáÉϽØÈ¡BP=AB£¬Á¬½ÓNP¡¢CP£¬

¡ßCN=$\frac{1}{2}$AB=2£¬CP=4$\sqrt{2}$£¬
¡àNP¡ÜCP-CN=4$\sqrt{2}$-2£¬
µ±C¡¢N¡¢PÈýµã¹²Ïßʱ£¬È¡µÃ×î´óÖµ£¬
ÓÖ¡ßMΪANµÄÖе㣬BΪAPµÄÖе㣬
¡àÏ߶ÎBMµÄ³¤¶ÈµÄ×îСֵΪBM=$\frac{1}{2}$NP¡Ü2$\sqrt{2}$-1£¬
ËùÒÔÏ߶ÎBMµÄ³¤¶ÈµÄ×îСֵΪ2$\sqrt{2}$-1£»
Èçͼ4£¬C¡¢N¡¢PÈýµã¹²Ïߣ¬

BE=4£¬EN=4$\sqrt{2}$-2£¬
ÉèN£¨x£¬y£©£¬$\frac{y}{BC}$=$\frac{EN}{EC}$£¬µÃ$\frac{y}{4}$=$\frac{4\sqrt{2}-2}{4\sqrt{2}}$£¬½âµÃy=4-$\sqrt{2}$£¬
$\frac{6-x}{4}$=$\frac{4\sqrt{2}-2}{4\sqrt{2}}$£¬½âµÃx=2+$\sqrt{2}$
¡à´ËʱN£¨2+$\sqrt{2}$£¬4-$\sqrt{2}$£©£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËÒ»´Îº¯ÊýµÄ×ÛºÏÌ⣬Éæ¼°Ò»´Îº¯Êý½âÎöʽ¡¢È«µÈÈý½ÇÐεÄÅж¨¡¢Èý½ÇÐεÄÈý±ß¹Øϵ¼°ÏàËÆÈý½ÇÐεĶÔÓ¦±ßµÄ±È£¬½âÌâµÄ¹Ø¼üÊǵ±C¡¢N¡¢PÈýµã¹²Ïßʱ£¬È¡µÃBMµÄ³¤¶ÈµÄ×îСֵ£®