{"id":6299,"date":"2021-10-12T16:48:59","date_gmt":"2021-10-12T11:18:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6299"},"modified":"2021-11-30T16:00:56","modified_gmt":"2021-11-30T10:30:56","slug":"normal-form-of-a-line-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/normal-form-of-a-line-equation\/","title":{"rendered":"Normal Form of a Line Equation"},"content":{"rendered":"

Here you will learn normal form of a line equation with proof and examples.<\/p>\n

Let’s begin –<\/p>\n

Normal Form of a Line (Perpendicular form of line)<\/h2>\n
\n

The equation of the straight line upon which the length of of the perpendicular from the origin is p and this perpendicular makes an angle \\(\\alpha\\) with x-axis is<\/p>\n

\\(xcos\\alpha\\) + \\(ysin\\alpha\\) = p.<\/p>\n<\/blockquote>\n

Proof<\/strong> : <\/p>\n

Let the line AB be such that the length of the perpendicular OQ from the origin O to the line be p and \\(\\angle\\) XOQ = \\(\\alpha\\)<\/p>\n

Let P(x, y) be any point on the line. Draw PL \\(\\perp\\) OX, LM \\(\\perp\\) OQ and PN \\(\\perp\\) LM. Then,\"normal<\/p>\n

OL = x and LP = y<\/p>\n

In triangle OLM, we have<\/p>\n

\\(cos\\alpha\\) = \\(OM\\over OL\\)<\/p>\n

\\(\\implies\\) OM = OL \\(cos\\alpha\\) = x \\(cos\\alpha\\).<\/p>\n

In triangle PNL, we have<\/p>\n

\\(sin\\alpha\\) = \\(PN\\over PL\\)<\/p>\n

\\(\\implies\\) PN = PL \\(sin\\alpha\\) = y \\(sin\\alpha\\)<\/p>\n

\\(\\implies\\) MQ = PN = y \\(sin\\alpha\\)<\/p>\n

Now, p = OQ = OM + MQ = \\(xcos\\alpha\\) + \\(ysin\\alpha\\)<\/p>\n

Hence, the equation of the required line is<\/p>\n

\\(xcos\\alpha\\) + \\(ysin\\alpha\\) = p. <\/p>\n

Example<\/strong><\/span> : Find the equation of the line which is at a distance 3 from the origin and the perpendicular from the origin to the line makes an angle of 30 with the positive direction of the x-axis.<\/p>\n

Solution<\/span><\/strong> : Here, p = 3, \\(\\alpha\\) = 30<\/p>\n

Equation of the line in the normal form is<\/p>\n

x cos 30 + y sin 30 = 3<\/p>\n

\\(\\implies\\) x \\(\\sqrt{3}\\over 2\\) + \\(y\\over 2\\) = 3<\/p>\n

\\(\\implies\\) \\(\\sqrt{3}\\) x + y = 6.<\/p>\n

which is the required equation line.<\/p>\n\n\n

\n
Next – Equation of a Line Perpendicular to a Line<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – Intercept Form of a Line Equation<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn normal form of a line equation with proof and examples. Let’s begin – Normal Form of a Line (Perpendicular form of line) The equation of the straight line upon which the length of of the perpendicular from the origin is p and this perpendicular makes an angle \\(\\alpha\\) with x-axis is …<\/p>\n

Normal Form of a Line Equation<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[32],"tags":[618,620,621,619],"yoast_head":"\nNormal Form of a Line Equation - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post you will learn normal form of a line equation with proof and examples.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/normal-form-of-a-line-equation\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Normal Form of a Line Equation - 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