ÌâÄ¿ÄÚÈÝ

Èçͼ£¬ÒÑÖª·´±ÈÀýº¯ÊýµÄͼÏñ¾­¹ýµÚ¶þÏóÏÞÄڵĵãA£¨£­1£¬m£©£¬AB¡ÍxÖáÓÚµãB£¬¡÷AOB µÄÃæ»ýΪ2£®ÈôÖ±Ïß y="ax+b" ¾­¹ýµãA£¬²¢ÇÒ¾­¹ý·´±ÈÀýº¯ÊýµÄͼÏóÉÏÁíÒ»µãC£¨n£¬Ò»2£©£®

£¨1£©Çó·´±ÈÀýº¯ÊýÓëÖ±Ïßy=ax+bµÄ½âÎöʽ£»

£¨2£©¸ù¾ÝËù¸øÌõ¼þ£¬Ö±½Óд³ö²»µÈʽ ax+b¡ÝµÄ½â¼¯_________________;

£¨3£©Çó³öÏß¶ÎOAµÄ³¤£¬²¢Ë¼¿¼£ºÔÚxÖáÉÏÊÇ·ñ´æÔÚÒ»µãP,ʹµÃ¡÷PAOÊǵÈÑüÈý½ÇÐΣ¬Èç¹û´æÔÚ£¬ÇëÇó³öPµÄ×ø±ê£¬Èç¹û²»´æÔÚ£¬Çë˵Ã÷ÀíÓÉ¡£

 

¡¾´ð°¸¡¿

£¨1£©

£¨2£©x¡Ü-1 »ò    0£¼x¡Ü2   £¨3£©£¨£¬0£©£¬£¨-£¬0£©£¬£¨-2,0£©£¬£¨-8.5£¬0£©

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©¡ßµãA£¨-1£¬m£©ÔÚµÚ¶þÏóÏÞÄÚ£¬¡àAB = m£¬OB = 1£¬¡à

¼´£º£¬½âµÃ£¬¡àA (-1,4)£¬                   1 ·Ö

¡ßµãA (-1,4)£¬ÔÚ·´±ÈÀýº¯ÊýµÄͼÏñÉÏ£¬¡à4 =£¬½âµÃ£¬

¡à·´±ÈÀýº¯ÊýΪ£¬                     2 ·Ö

ÓÖ¡ß·´±ÈÀýº¯ÊýµÄͼÏñ¾­¹ýC£¨n£¬£©¡à£¬½âµÃ£¬¡àC (2,-2)£¬

¡ßÖ±Ïß¹ýµãA (-1,4)£¬C (2,-2)

¡à  ½â·½³Ì×éµÃ 

¡àÖ±ÏߵĽâÎöʽΪ £»              4·Ö

(2) x¡Ü-1 »ò    0£¼x¡Ü2                                     6·Ö

£¨3£©´æÔÚ¡£PµÄ×ø±êΪ£º£¨£¬0£©£¬£¨-£¬0£©£¬£¨-2,0£©£¬£¨-8.5£¬0£©

£¨Ã¿¸öµã¸÷1·Ö£©              

¿¼µã£ºÒ»´Îº¯Êý£¬·´±ÈÀýº¯Êý

µãÆÀ£º±¾ÌâÊôÓÚ¶ÔÒ»´Îº¯ÊýºÍ·´±ÈÀýº¯ÊýµÄ»ù±¾ÖªÊ¶µÄÀí½âºÍÔËÓÃ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø