A perpendicular line is a straight line through a point. It makes an angle of 90 degrees with a particular point through which the line passes. Coordinates and line equation is the prerequisite to finding out the perpendicular line.
Consider the equation of the line is ax + by + c = 0 and coordinates are (x1, y1), the slope should be −a/b. If one line is perpendicular to this line, the product of slopes should be -1. Let m1 and m2 be the slopes of two lines, and if they are perpendicular to each other, then their product will be -1.
\[\large Perpendicular\;Lines;\;m_{1}\times m_{2}=-1\]
\[\large Slope\;m=\frac{-a}{b}\]
\[\large Perpendicular\;Line \; equation:\; (y-y_{1})=m(x-x_{1})\]
Solved Example
Question: Check whether 2x+ 3y + 5 = 0 and 3x – 2y + 1 = 0 are perpendicular or not.
Solution:
The given equations of lines are:
2x + 3y + 5 = 0 and 3x– 2y + 1 = 0
To check whether they are perpendicular to each other, find out the slopes of both lines. If the product of their slopes is -1, these lines are perpendicular to each other.
Slope formula is; m = \(\begin{array}{l}\frac{-a}{b}\end{array} \)
Slope for first line,
\(\begin{array}{l}m_{1}\end{array} \)
= \(\begin{array}{l}\frac{-a}{b}\end{array} \) \(\begin{array}{l}\frac{-2}{3}\end{array} \) Slope for second line, \(\begin{array}{l}m_{2}\end{array} \) \(\begin{array}{l}\frac{-a}{b}\end{array} \) \(\begin{array}{l}\frac{-3}{-2}\end{array} \) \(\begin{array}{l}m_{1}\times m_{2}\end{array} \) \(\begin{array}{l}\frac{-2}{3}\end{array} \) \(\begin{array}{l}\times\end{array} \) \(\begin{array}{l}\frac{-3}{-2}\end{array} \) Since the product of slope is -1, the given lines are perpendicular to each other.
