<\/ins><\/p>\n\n\n\na) <\/p>\n\n\n\n
+ A (4; 3) thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng y = ax + b \u21d2 3 = 4.a + b (1)<\/p>\n\n\n\n
+ B (2; \u20131) thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng y = ax + b \u21d2 \u20131 = 2.a + b (2)<\/p>\n\n\n\n
L\u1ea5y (1) tr\u1eeb (2) ta \u0111\u01b0\u1ee3c: 3 \u2013 (\u20131) = (4a + b) \u2013 (2a + b) <\/p>\n\n\n\n
\u21d2 4 = 2a \u21d2 a = 2 \u21d2 b = \u20135. <\/p>\n\n\n\n
V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m A(4;3), B(2 ; \u20131) l\u00e0 y = 2x \u2013 5. <\/p>\n\n\n\n
b) <\/p>\n\n\n\n
+ \u0110\u01b0\u1eddng th\u1eb3ng song song v\u1edbi Ox c\u00f3 d\u1ea1ng y = b.<\/p>\n\n\n\n
+ \u0110\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m A(1 ; \u20131) n\u00ean b = \u2013 1. <\/p>\n\n\n\n
V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng c\u1ea7n t\u00ecm l\u00e0 y = \u20131. <\/p>\n\n\n\n
Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ \u0110\u01b0\u1eddng th\u1eb3ng y = b (b l\u00e0 h\u1eb1ng s\u1ed1) l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c ho\u00e0nh. <\/p>\n\n\n\n
B\u00e0i 4 (trang 42 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1<\/p>\n\n\n\n