Here you will learn what is the equation of a line perpendicular to a line with proof and examples.
Let’s begin –
Equation of a Line Perpendicular to a Line
The equation of a line perpendicular to a given line ax + by + c = 0 is
bx – ay + \(\lambda\) = 0,
where \(\lambda\) is constant.
Let \(m_1\) be the slope of the given line and \(m_2\) be the slope of a line perpendicular to the given line. Then,
\(m_1\) = -\(a\over b\)
But, \(m_1m_2\) = -1 for perpendicular lines
\(\implies\) \(m_2\) = -\(1\over m_1\) = \(b\over a\)
Let \(c_2\) be the y-intercept of the required line. Then, its equation is
y = \(m_2\)x + \(c_2\)
y = \(b\over a\)x + \(c_2\)
\(\implies\) bx – ay + a\(c_2\) = 0
\(\implies\) bx – ay + \(\lambda\), where \(\lambda\) = a\(c_2\) = constant.
To write a line perpendicular to a given line we proceed as follows :
1). Interchanging x and y.
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.
3). Replace the given constant by a new constant \(\lambda\) which is determined by the given condition.
Example : Find the equation of the straight line that passes through the point (3, 4) and perpendicular to the line 3x + 2y + 5 = 0.
Solution : The line perpendicular to 3x + 2y + 5 = 0 is
2x – 3y + \(\lambda\) = 0 ……………(i)
This passes through the point (3, 4)
\(\therefore\) 3 \(\times\) 2 – 3 \(\times\) 4 + \(\lambda\) = 0 \(\implies\) \(\lambda\) = 6
Putting \(\lambda\) = 6 in equation (i), we get
2x – 3y + 6 = 0, which is the required equation.