{"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
\nThe 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,<\/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