{"id":4930,"date":"2012-11-16T23:21:28","date_gmt":"2012-11-16T23:21:28","guid":{"rendered":"http:\/\/SchoolTutoring.com\/help\/?p=4930"},"modified":"2014-12-02T08:31:56","modified_gmt":"2014-12-02T08:31:56","slug":"equation-of-a-line-normal-form","status":"publish","type":"post","link":"https:\/\/schooltutoring.com\/help\/equation-of-a-line-normal-form\/","title":{"rendered":"Equation of a Line &#8211; Normal Form"},"content":{"rendered":"<p><strong>Normal:<br \/>\n<\/strong><\/p>\n<p>A normal to a line is a line segment drawn from a point perpendicular to the given line.<\/p>\n<p><a href=\"https:\/\/SchoolTutoring.com\/wp-content\/uploads\/sites\/2\/2012\/11\/distance1.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4931\" title=\"distance1\" src=\"https:\/\/SchoolTutoring.com\/wp-content\/uploads\/sites\/2\/2012\/11\/distance1.jpg\" alt=\"\" width=\"203\" height=\"113\" \/><\/a><\/p>\n<p>Let <strong><em>p<\/em><\/strong> be the length of the normal drawn from the origin to a line, which subtends an angle \u00f8 with the positive direction of x-axis as follows.<\/p>\n<p><a href=\"https:\/\/SchoolTutoring.com\/wp-content\/uploads\/sites\/2\/2012\/11\/distance2.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4932\" title=\"distance2\" src=\"https:\/\/SchoolTutoring.com\/wp-content\/uploads\/sites\/2\/2012\/11\/distance2.jpg\" alt=\"\" width=\"305\" height=\"211\" srcset=\"https:\/\/schooltutoring.com\/help\/wp-content\/uploads\/sites\/2\/2012\/11\/distance2.jpg 305w, https:\/\/schooltutoring.com\/help\/wp-content\/uploads\/sites\/2\/2012\/11\/distance2-300x207.jpg 300w\" sizes=\"auto, (max-width: 305px) 100vw, 305px\" \/><\/a><\/p>\n<p>Then, we have Cos \u00f8 = p\/m \u00e0 m = p\/Cos \u00f8<\/p>\n<p>And Sin \u00f8 = p\/n \u00e0 n = p\/Sin \u00f8<\/p>\n<p>The equation of line in intercept form is,<\/p>\n<p>x\/m + y\/n =1<\/p>\n<p>x\/(p\/Cos \u00f8) + y\/(p\/Sin \u00f8) = 1<\/p>\n<p>x Cos \u00f8\/p + y Sin \u00f8\/p =1<\/p>\n<p><strong><em>x Cos \u00f8 + y Sin \u00f8 = p.<\/em><\/strong><\/p>\n<p>This is called the normal form of equation of the given line making the angle \u00f8 with the positive direction of x-axis and whose perpendicular distance from the origin is <strong><em>p<\/em><\/strong>.<\/p>\n<p><strong>Converting the general equation of a line into normal form:<\/strong><\/p>\n<p>The equation of a straight line in general form is,<\/p>\n<p>ax+by+c=0<\/p>\n<p>From the above equations, we have<\/p>\n<p>Cos \u00f8 = p\/m and Sin \u00f8 = p\/n<\/p>\n<p>From Trigonometric identity, we have<\/p>\n<p>Cos<sup>2 <\/sup>\u00f8 + Sin<sup>2<\/sup>\u00f8=1<\/p>\n<p>p<sup>2<\/sup>\/m<sup>2<\/sup> + p<sup>2<\/sup>\/n<sup>2<\/sup>=1<\/p>\n<p>Solving this equation gives us,<\/p>\n<p>P = mn\/\u221a(m<sup>2<\/sup>+n<sup>2<\/sup>), which is the perpendicular distance from the origin to the line x\/m + y\/n =1 (i.e., nx + my \u2013 mn=0)<\/p>\n<p>Thus, we have the distance from origin to the line ax+by+c=0, = |c|\/\u221a(a<sup>2<\/sup>+b<sup>2<\/sup>)<\/p>\n<p>Thus, for converting the given line into normal form, divide the equation ax+by+c=0 by \u221a(a<sup>2<\/sup>+b<sup>2<\/sup>).<br \/>\nThus,<\/p>\n<p><strong><em>[a\/ \u221a(a<sup>2<\/sup>+b<sup>2<\/sup>)] x + [b\/\u221a(a<sup>2<\/sup>+b<sup>2<\/sup>)] + c\/\u221a(a<sup>2<\/sup>+b<sup>2<\/sup>)=0<\/em><\/strong><\/p>\n<p>We offer Study Skills tutoring, <a href=\"https:\/\/schooltutoring.com\/tutoring-programs\/study-skills-workshops\/\">click here<\/a> for more information.<\/p>\n<p><span class=\"tutorOrange\">SchoolTutoring Academy<\/span> is the premier educational services company for K-12 and college students. We offer tutoring programs for students in K-12, AP classes, and college. To learn more about how we help parents and students in North-Dakota visit: <a href=\"https:\/\/schooltutoring.com\/North-Dakota-Tutoring-Programs\/\">Tutoring in North-Dakota. <\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Normal: A normal to a line is a line segment drawn from a point perpendicular to the given line. Let p be the length of the normal drawn from the origin to a line, which subtends an angle \u00f8 with the positive direction of x-axis as follows. Then, we have Cos \u00f8 = p\/m \u00e0 [&hellip;]<\/p>\n","protected":false},"author":19,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"categories":[10],"tags":[94,398,515,721,1201,1265,1319,1859],"class_list":["post-4930","post","type-post","status-publish","format-standard","hentry","category-geometry","tag-angle","tag-converting","tag-distance","tag-general-equation","tag-normal","tag-origin","tag-perpendicular","tag-trigonometric"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/4930","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/users\/19"}],"replies":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/comments?post=4930"}],"version-history":[{"count":0,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/4930\/revisions"}],"wp:attachment":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/media?parent=4930"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/categories?post=4930"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/tags?post=4930"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}