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HomeLớp 10Toán lớp 10 - Sách Kết nối tri thứcBài 20. Đường thẳng trong mặt phẳng tọa độBài tập cơ bản
{"common":{"save":0,"post_id":"6475","level":1,"total":10,"point":10,"point_extra":0},"segment":[{"id":"7072","mon_id":"1158767","chapter_id":"1158892","post_id":"6475","question":"<p>Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxy, cho hai \u0111\u01b0\u1eddng th\u1eb3ng <span class=\"math-tex\">$\\Delta_1:a_1x+b_1y+c_1=0$<\/span> và <span class=\"math-tex\">$\\Delta_2:a_2x+b_2y+c_2=0$<\/span>. <\/p><p>Cosin c\u1ee7a góc gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng \u0111ó \u0111\u01b0\u1ee3c xác \u0111\u1ecbnh theo công th\u1ee9c<\/p>","options":["<strong>A.<\/strong> cos<span class=\"math-tex\">$\\big(\\Delta_1,\\Delta_2\\big)=\\dfrac{a_1.a_2+b_1b_2}{\\sqrt{a_1^2+b_1^2}.\\sqrt{a_2^2+b_2^2}}$<\/span>","<strong>B.<\/strong> cos<span class=\"math-tex\">$\\big(\\Delta_1,\\Delta_2\\big)=\\dfrac{|a_1.a_2+b_1b_2|}{\\sqrt{a_1^2+b_1^2}.\\sqrt{a_2^2+b_2^2}}$<\/span>","<strong>C.<\/strong> cos<span class=\"math-tex\">$\\big(\\Delta_1,\\Delta_2\\big)=\\dfrac{|a_1.a_2+b_1b_2|}{\\sqrt{a_1^2+b_1^2}+\\sqrt{a_2^2+b_2^2}}$<\/span>","<strong>D.<\/strong> cos<span class=\"math-tex\">$\\big(\\Delta_1,\\Delta_2\\big)=\\sqrt{\\dfrac{a_1.a_2+b_1b_2+c_1c_2}{a^2+b^2}}$<\/span>"],"correct":"2","level":"1","hint":"","answer":"<p>\u0110áp án \u0111úng là <span style=\"color:#16a085;\"> <strong>B.<\/strong> cos<span class=\"math-tex\">$\\big(\\Delta_1,\\Delta_2\\big)=\\dfrac{|a_1.a_2+b_1b_2|}{\\sqrt{a_1^2+b_1^2}.\\sqrt{a_2^2+b_2^2}}$<\/span> .<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-06-01 09:55:37","option_type":"math","len":0},{"id":"7074","mon_id":"1158767","chapter_id":"1158892","post_id":"6475","question":"<p>Trong m\u1eb7t ph\u1eb3ng h\u1ec7 to\u1ea1 \u0111\u1ed9 Oxy, cho \u0111i\u1ec3m M(<span class=\"math-tex\">$x_0;y_0$<\/span>)<\/p><p>và \u0111\u01b0\u1eddng th\u1eb3ng <span class=\"math-tex\">$\\Delta$<\/span>: ax + by + c = 0 v\u1edbi a² + b² > 0.<\/p><p>Kho\u1ea3ng cách t\u1eeb \u0111i\u1ec3m M \u0111\u1ebfn <span class=\"math-tex\">$\\Delta$<\/span> \u0111\u01b0\u1ee3c tính theo công th\u1ee9c<\/p>","options":["<strong>A.<\/strong> d(M; <span class=\"math-tex\">$\\Delta$<\/span>) = <span class=\"math-tex\">$\\dfrac{|ax_0+by_0|}{a^2+b^2}$<\/span> .","<strong>B.<\/strong> d(M; <span class=\"math-tex\">$\\Delta$<\/span>) = <span class=\"math-tex\">$\\dfrac{ax_0+by_0}{a^2+b^2}$<\/span> .","<strong>C.<\/strong> d(M; <span class=\"math-tex\">$\\Delta$<\/span>) = <span class=\"math-tex\">$\\dfrac{|ax_0+by_0+c|}{\\sqrt{a^2+b^2}}$<\/span> .","<strong>D.<\/strong> d(M; <span class=\"math-tex\">$\\Delta$<\/span>) = <span class=\"math-tex\">$\\dfrac{ax_0+by_0+c}{\\sqrt{a^2+b^2}}$<\/span> ."],"correct":"3","level":"1","hint":"","answer":"<p>\u0110áp án \u0111úng là <span style=\"color:#16a085;\"> <strong>C.<\/strong> d(M; <span class=\"math-tex\">$\\Delta$<\/span>) = <span class=\"math-tex\">$\\dfrac{|ax_0+by_0+c|}{\\sqrt{a^2+b^2}}$<\/span> .<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-06-01 10:01:36","option_type":"math","len":0},{"id":"7075","mon_id":"1158767","chapter_id":"1158892","post_id":"6475","question":"<p>Xét v\u1ecb trí t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng a: 2x + y – 1 = 0 và b: x – 2y = 0.<\/p>","options":["<strong>A.<\/strong> Trùng nhau","<strong>B.<\/strong> Song song","<strong>C.<\/strong> Vuông góc v\u1edbi nhau","<strong>D.<\/strong> C\u1eaft nhau nh\u01b0ng không vuông góc"],"correct":"3","level":"1","hint":"","answer":"<p>a có vect\u01a1 pháp tuy\u1ebfn <span class=\"math-tex\">$\\overrightarrow{u}$<\/span> = (2; 1).<\/p><p>b có vect\u01a1 pháp tuy\u1ebfn <span class=\"math-tex\">$\\overrightarrow{v}$<\/span> = (1; –2).<\/p><p>Ta có <span class=\"math-tex\">$\\overrightarrow{u}.\\overrightarrow{v}$<\/span> = 2.1 + 1.(–2) = 0.<\/p><p>V\u1eady a và b vuông góc v\u1edbi nhau.<\/p><p>\u0110áp án \u0111úng là <span style=\"color:#16a085;\"><strong>C.<\/strong> Vuông góc v\u1edbi nhau.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-06-01 10:05:59","option_type":"txt","len":3}]}