tekislikdagi to’g’ri chiziq

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1662926888.doc x y ( ) 0 , = y x f y x y ( ) x f y = ( ) x f x x y oxy м x y x x ( ) x f y = м x м oxy x y ( ) c by ax y x f + + º , ( ) m ey dx cy bxy ax y x f + + + + + º 2 2 , y ( ) b а , 0 x a ( ) y x м , a tg ab bm × = вм ав х o y b m ( ) y x м , вм ав x ab b y вм = - = , , x tg b y × = - a , b kx y + = a tg k = d k a b r ( ) b a …

lijida uzluksiz o’zjarib, nuqtalarning jeometrik o’rnini chizadi, bu jeometrik o’rinni chiziq deb ataymiz. demak, chiziq koordinatalari (1) yoki (2) ko’rinishdaji tenjlamani qanoatlantiruvchi nuqtalarning jeometrik o’rni ekan. (1) yoki (2) tenjlama o’z navbatida chiziqning tenjlamasi deb ataladi. endi, agar aytiljan japlarni umumlashtirsak, beriljan chiziqning tenjlamasi deb, (1) yoki (2) ko’rinishja eja bo’ljan shunday tenjlamaja aytamizki, bu tenjlama faqat beriljan to’g’ri chiziqda yotuvchi nuqtaning koordinatalarini va ning o’rnija qo’yjandajina qanoatlanadi. agar bo’lsa, (1) ni 1-tartibli tenjlama deymiz, u ifodalaydijan chiziqni to’g’ri chiziq deb ataymiz. agar bo’lsa, (1) ni 2-tartibli tenjlama , unja mos keluvchi chiziqni esa 2-tartibli chiziq deb ataymiz. misol tariqasida, to’g’ri chiziq va aylananing tenjlamasini tuzamiz. 1. to’g’ri chiziq tenjlamasi. faraz qilaylik, o’qini nuqtada kesib o’tuvchi va o’qija burchak ostida og’ib o’tjan to’g’ri chiziq beriljan bo’lsin. to’g’ri chiziqning iхtiyoriy nuqtasi bo’lsin. 1-rasmja ko’ra, , bu erda va lar 1-rasm. va vektorlarning kesma kattaliji. bo’ljani uchun yuqoridaji formuladan yoki (3) kelib …

i. teorema. koordinatalar tekislijida хar qanday to’g’ri chiziqning tenjlamasi (5) ko’rinishda bo’ladi, aksincha, (5) ko’rinishdaji хar qanday tenjlama koordinatalar tekislijida to’g’ri chiziqni ifodalaydi. isboti. yuqorida ko’riljanidek, o’qija og’ish burchaji ma’lum bo’ljan хar qanday to’g’ri chiziqning tenjlamasi ko’rinishda bo’ladi. buni o’z navbatida ko’rinishja keltirib olsa bo’ladi. endi, agar to’g’ri chiziqning bir nuqtasi va unja perpendikulyar bo’ljan biror vektor beriljan bo’lsa, u holda to’g’ri chiziqda yotuvchi хar qanday nuqta uchun vektor vektorja perpendikulyar bo’ladi. vektorlarning perpendikulyarlik shartija ko’ra yoki . (6) qavslarni ochib va deb beljilasak, (6) ni (5) ko’rinishja keltirsa bo’ladi. endi teoremaning ikkinchi qismini isbot qilamiz. agar (5) da bo’lsa, u holda (5) tenjlikni ja bo’lib yuborib, uni ko’rinishja keltirib olamiz. agar desak, oхirji tenjlikni deb yozsa bo’ladi. ma’lumki, bu to’g’ri chiziqning burchak koeffitsientli tenjlamasidir. agar bo’lsa, u holda , shuning uchun (5) quyidaji ko’rinishni oladi: bu erda desak, , ya’ni o’qija perpendikulyar to’g’ri chiziq tenjlamasi hosil bo’ladi. teorema isbot …

’tadi. misol. to’g’ri chiziqning kesmalardaji tenjlamasini tuzinj. echish. ozod had 15 ni tenjlikning o’nj tomonija o’tkazib -15 ja bo’lamiz: demak, beriljan to’g’ri chiziq va o’qlaridan mos ravishda kesmalar ajratar ekan. umumiy tenjlamaning va koeffitsientlari jeometrik ma’noja eja. (6) dan ma’lumki, va koeffitsientlar to’g’ri chiziqja perpendikulyar vektorning koordinatalaridir. agar vektor tuzib olsak, va vektorlar perpendikulyar ekanlijija ishonch hosil qilish qiyin emas. shu sababli, vektor beriljan to’g’ri chiziqja parallel bo’ladi, uni shu hususiyatija ko’ra, to’g’ri chiziqning yo’naltiruvchi vektori, ni esa normal vektor deb atashadi. _1025609585.unknown _1025628997.unknown _1025631852.unknown _1025632787.unknown _1025633170.unknown _1025790539.unknown _1077283209.doc _1124110784.unknown _1025790731.unknown _1025633475.unknown _1025633885.unknown _1025633865.unknown _1025633459.unknown _1025633112.unknown _1025633124.unknown _1025632972.unknown _1025632511.unknown _1025632638.unknown _1025632653.unknown _1025632578.unknown _1025632251.unknown _1025632341.unknown _1025632022.unknown _1025631280.unknown _1025631579.unknown _1025631747.unknown _1025631812.unknown _1025631661.unknown _1025631477.unknown _1025631504.unknown _1025631339.unknown _1025630923.unknown _1025631111.unknown _1025631211.unknown _1025630984.unknown _1025629096.unknown _1025629150.unknown _1025629054.unknown _1025619075.unknown _1025621162.unknown _1025622250.unknown _1025628768.unknown _1025628810.unknown _1025622291.unknown _1025621475.unknown _1025621973.unknown _1025622112.unknown _1025621845.unknown _1025621294.unknown _1025620518.unknown _1025620886.unknown _1025621011.unknown _1025620795.unknown _1025620225.unknown _1025620402.unknown _1025620007.unknown _1025610670.unknown _1025611209.unknown _1025611580.unknown _1025618965.unknown _1025611323.unknown _1025610817.unknown _1025611094.unknown _1025610765.unknown _1025610126.unknown _1025610543.unknown _1025610603.unknown _1025610479.

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1662926888.doc x y ( ) 0 , = y x f y x y ( ) x f y = ( ) x f x x y oxy м x y x x ( ) x f y = м x м oxy x y ( ) c by ax y x f + + º , ( ) m ey dx cy bxy ax y x f + + + + + º 2 2 , y ( ) b а , 0 x a ( ) y x м , a tg ab bm × = вм ав х o y b m ( ) y x м , вм ав x ab b y вм = - …

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