{"id":6301,"date":"2021-10-12T16:50:36","date_gmt":"2021-10-12T11:20:36","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6301"},"modified":"2021-10-12T22:49:59","modified_gmt":"2021-10-12T17:19:59","slug":"equation-of-a-line-perpendicular-to-a-line","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-a-line-perpendicular-to-a-line\/","title":{"rendered":"Equation of a Line Perpendicular to a Line"},"content":{"rendered":"
Here you will learn what is the equation of a line perpendicular to a line with proof and examples.<\/p>\n
Let’s begin –<\/p>\n
\nThe equation of a line perpendicular to a given line ax + by + c = 0 is<\/p>\n
bx – ay + \\(\\lambda\\) = 0,<\/p>\n
where \\(\\lambda\\) is constant.<\/p>\n<\/blockquote>\n
Proof :<\/strong><\/p>\n
\nLet \\(m_1\\) be the slope of the given line and \\(m_2\\) be the slope of a line perpendicular to the given line. Then,<\/p>\n
\\(m_1\\) = -\\(a\\over b\\)<\/p>\n
But, \\(m_1m_2\\) = -1 for perpendicular lines<\/p>\n
\\(\\implies\\) \\(m_2\\) = -\\(1\\over m_1\\) = \\(b\\over a\\)<\/p>\n
Let \\(c_2\\) be the y-intercept of the required line. Then, its equation is<\/p>\n
y = \\(m_2\\)x + \\(c_2\\)<\/p>\n
y = \\(b\\over a\\)x + \\(c_2\\)<\/p>\n
\\(\\implies\\) bx – ay + a\\(c_2\\) = 0<\/p>\n
\\(\\implies\\) bx – ay + \\(\\lambda\\), where \\(\\lambda\\) = a\\(c_2\\) = constant.<\/p>\n<\/blockquote>\n
To write a line perpendicular to a given line we proceed as follows<\/strong> :<\/p>\n
\n1). Interchanging x and y.<\/p>\n
2). If the coefficients of x and y in the given equation are of the same sign make them of opposite sign and if the coefficients are of opposite signs make them of the same sign.<\/p>\n
3). Replace the given constant by a new constant \\(\\lambda\\) which is determined by the given condition.<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Find the equation of the straight line that passes through the point (3, 4) and perpendicular to the line 3x + 2y + 5 = 0.<\/p>\n
Solution<\/strong><\/span> : The line perpendicular to 3x + 2y + 5 = 0 is<\/p>\n
2x – 3y + \\(\\lambda\\) = 0 ……………(i)<\/p>\n
This passes through the point (3, 4)<\/p>\n
\\(\\therefore\\) 3 \\(\\times\\) 2 – 3 \\(\\times\\) 4 + \\(\\lambda\\) = 0 \\(\\implies\\) \\(\\lambda\\) = 6 <\/p>\n
Putting \\(\\lambda\\) = 6 in equation (i), we get<\/p>\n
2x – 3y + 6 = 0, which is the required equation.<\/p>\n\n\n