Burchak koeffitsiyenti tenglamasi berilgan toʻgʻri chiziqni chizish

Agar hali oʻqimagan boʻlsangiz, mana bundan boshlashingiz mumkin: Chiziqli funksiyaning umumiy koʻrinishi mavzusiga kirish.

Keling, $y=2x+3$y, equals, 2, x, plus, 3 ni chizamiz.

Eslatib oʻtamiz, umumiy chiziqli funksiya tenglamasi $y=\maroonC{m}x+\greenE{b}$y, equals, start color #ed5fa6, m, end color #ed5fa6, x, plus, start color #0d923f, b, end color #0d923f boʻlib, bu yerda burchak koeffitsiyenti $\maroonC{m}$start color #ed5fa6, m, end color #ed5fa6 hamda $y$y-kesishish nuqtasi $\greenE{b}$start color #0d923f, b, end color #0d923f ga teng edi. Shunga koʻra $y=\maroonC{2}x+\greenE{3}$y, equals, start color #ed5fa6, 2, end color #ed5fa6, x, plus, start color #0d923f, 3, end color #0d923f tenglamada burchak koeffitsiyenti $\maroonC{2}$start color #ed5fa6, 2, end color #ed5fa6 va $y$y-kesishish nuqtasi $(0,\greenE{3})$left parenthesis, 0, comma, start color #0d923f, 3, end color #0d923f, right parenthesis ga teng.

Chiziqni grafikda koʻrsatish uchun mazkur chiziqda yotuvchi ikkita nuqta kerak boʻladi. Biz $(0,\greenE{3})$left parenthesis, 0, comma, start color #0d923f, 3, end color #0d923f, right parenthesis nuqtaning chiziqda joylashganini bilamiz.

Shu bilan birga, chiziqning burchak koeffitsiyenti $\maroonC{2}$start color #ed5fa6, 2, end color #ed5fa6 boʻlgach, $(0\maroonC{+1},\greenE{3}\maroonC{+2})=(1,5)$left parenthesis, 0, start color #ed5fa6, plus, 1, end color #ed5fa6, comma, start color #0d923f, 3, end color #0d923f, start color #ed5fa6, plus, 2, end color #ed5fa6, right parenthesis, equals, left parenthesis, 1, comma, 5, right parenthesis nuqta ham chiziqda yotadi.

Keling, $y=\maroonC{\dfrac{2}{3}}x\greenE{+1}$y, equals, start color #ed5fa6, start fraction, 2, divided by, 3, end fraction, end color #ed5fa6, x, start color #0d923f, plus, 1, end color #0d923f ning grafigini chizamiz.

Oldingi kabi chiziq quyidagi nuqtalar orqali oʻtadi: $y$y-kesishish nuqtasi $(0,\greenE{1})$left parenthesis, 0, comma, start color #0d923f, 1, end color #0d923f, right parenthesis va qoʻshimcha nuqta $\left(0\maroonC{+1},\greenE{1}\maroonC{+\dfrac{2}{3}}\right)=\left(1,1\dfrac{2}{3}\right)$left parenthesis, 0, start color #ed5fa6, plus, 1, end color #ed5fa6, comma, start color #0d923f, 1, end color #0d923f, start color #ed5fa6, plus, start fraction, 2, divided by, 3, end fraction, end color #ed5fa6, right parenthesis, equals, left parenthesis, 1, comma, 1, start fraction, 2, divided by, 3, end fraction, right parenthesis.

$\left(1,1\dfrac{2}{3}\right)$left parenthesis, 1, comma, 1, start fraction, 2, divided by, 3, end fraction, right parenthesis nuqta chiziqda yotishi haqiqat ekan, kasrli koordinatalar bilan butun sonli koordinatar singari aniq nuqtalarni belgilay olmaymiz.

Chiziqdan koordinatalari butun son boʻlgan boshqa nuqta topish kerak. Buning uchun burchak koeffitsiyenti $\maroonC{\dfrac{2}{3}}$start color #ed5fa6, start fraction, 2, divided by, 3, end fraction, end color #ed5fa6 boʻlganda $x$x ning $\maroonC{3}$start color #ed5fa6, 3, end color #ed5fa6 birlik oʻsishi $y$y ning $\maroonC{2}$start color #ed5fa6, 2, end color #ed5fa6 birlik oʻsishiga sabab boʻladi degan dalildan foydalanamiz.

Bu bizga qoʻshimcha nuqtalar beradi: $(0\maroonC{+3},\greenE{1}\maroonC{+2})=(3,3)$left parenthesis, 0, start color #ed5fa6, plus, 3, end color #ed5fa6, comma, start color #0d923f, 1, end color #0d923f, start color #ed5fa6, plus, 2, end color #ed5fa6, right parenthesis, equals, left parenthesis, 3, comma, 3, right parenthesis.