We will learn how to find the equation of a line perpendicular to a line.
Prove that the equation of a line perpendicular to a given line ax + by + c = 0 is bx  ay + λ = 0, where λ is a constant.
Let m\(_{1}\) be the slope of the given line ax + by + c = 0 and m\(_{2}\) be the slope of a line perpendicular to the given line.
Then,
m\(_{1}\) = \(\frac{a}{b}\) and m\(_{1}\)m\(_{2}\) = 1
⇒ m\(_{2}\) = \(\frac{1}{m_{1}}\) = \(\frac{b}{a}\)
Let c\(_{2}\) be the yintercept of the required line. Then its equation is
y = m\(_{2}\)x + c\(_{2}\)
⇒ y = \(\frac{b}{a}\) x + c\(_{2}\)
⇒ bx  ay + ac\(_{2}\) = 0
⇒ bx  ay + λ = 0, where λ = ac\(_{2}\) = constant.
To get it more clear let us assume that ax + by + c = 0 (b ≠ 0) be the equation of the given straight line.
Now convert the ax + by + c = 0 in to slopeintercept form we get,
by =  ax  c
⇒ y =  \(\frac{a}{b}\) x  \(\frac{c}{b}\)
Therefore, the slope of the straight line ax + by + c = 0 is ( \(\frac{a}{b}\)).
Let m be the slope of a line which is perpendicular to the line ax + by + c = 0. Then, we must have,
m × ( \(\frac{a}{b}\)) =  1
⇒ m = \(\frac{b}{a}\)
Therefore, the equation of a line perpendicular to the line ax + by + c = 0 is
y = mx + c
⇒ y = \(\frac{b}{a}\) x + c
⇒ ay = bx + ac
⇒ bx  ay+ k = 0, where k = ac, is an arbitrary constant.
Algorithm for directly writing the equation of a straight line perpendicular to a given straight line:
To write a straight line perpendicular to a given straight line we proceed as follows:
Step I: Interchange the coefficients of x and y in equation ax + by + c = 0.
Step II: Alter the sign between the terms in x and y of equation i.e., If the coefficient of x and y in the given equation are of the same signs make them of opposite signs and if the coefficient of x and y in the given equation are of the opposite signs make them of the same sign.
Step III: Replace the given constant of equation ax + by + c = 0 by an arbitrary constant.
For example, the equation of a line perpendicular to the line 7x + 2y + 5 = 0 is 2x  7y + c = 0; again, the equation of a line, perpendicular to the line 9x  3y = 1 is 3x + 9y + k = 0.
Note:
Assigning different values to k in bx  ay + k = 0 we shall get different straight lines each of which is perpendicular to the line ax + by + c = 0. Thus we can have a family of straight lines perpendicular to a given straight line.
Solved examples to find the equations of straight lines perpendicular to a given straight line
1. Find the equation of a straight line that passes through the point (2, 3) and perpendicular to the straight line 2x + 4y + 7 = 0.
Solution:
The equation of a line perpendicular to 2x + 4y + 7 = 0 is
4x  2y + k = 0 …………………… (i) Where k is an arbitrary constant.
According to the problem equation of the perpendicular line 4x  2y + k = 0 passes through the point (2, 3)
Then,
4 ∙ (2)  2 ∙ (3) + k = 0
⇒ 8  6 + k = 0
⇒  14 + k = 0
⇒ k = 14
Now putting the value of k = 14in (i) we get, 4x  2y + 14 = 0
Therefore the required equation is 4x  2y + 14 = 0.
2. Find the equation of the straight line which passes through the point of intersection of the straight lines x + y + 9 = 0 and 3x  2y + 2 = 0 and is perpendicular to the line 4x + 5y + 1 = 0.
Solution:
The given two equations are x + y + 9 = 0 …………………… (i) and 3x  2y + 2 = 0 …………………… (ii)
Multiplying equation (i) by 2 and equation (ii) by 1 we get
2x + 2y + 18 = 0
3x  2y + 2 = 0
Adding the above two equations we get, 5x =  20
⇒ x =  4
Putting x = 4 in (i) we get, y = 5
Therefore, the coordinates of the point of intersection of the lines (i) and (ii) are ( 4,  5).
Since the required straight line is perpendicular to the line 4x + 5y + 1 = 0, hence we assume the equation of the required line as
5x  4y + λ = 0 …………………… (iii)
Where λ is an arbitrary constant.
By problem, the line (iii) passes through the point ( 4,  5); hence we must have,
⇒ 5 ∙ ( 4)  4 ∙ ( 5) + λ = 0
⇒ 20 + 20 + λ = 0
⇒ λ = 0.
Therefore, the equation of the required straight line is 5x  4y = 0.
● The Straight Line
11 and 12 Grade Math
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