What is the easiest way to find the line perpendicular to a given line?

Asked on by brandyboy

2 Answers | Add Yours

hustoncmk's profile pic

hustoncmk | High School Teacher | (Level 1) Assistant Educator

Posted on

Lines that are perpendicular have slopes that are negative reciprocals of each other.  What this means is if the slope for the first line is a/b the perpendicular line would have a slope of -b/a.

Let's start with an example with numbers so that you can see the full process.

Find the line perpendicular to y = 1/3x + 2, that crosses at point (3,2)

It is easiest to start with a line in slope-intercept form. For our example we will use y = 1/3x + 2 as the original line. 1/3 is the slope so we know that the line goes up 1 unit and right 3 units. The 2 is the y-intercept, this tells us that this line crosses the y axis at (0,2).

To find the perpendicular line, the first step is to find the negative reciprical of the slope 1/3.  Invert the fraction to 3/1 and add a negative sign.  The new slope is -3. One point is given as (3,2). If you substitute these into the equation with the new slope you can solve for the slope intercept and complete the new equation.

y = -3x + b

2 = -3(3) + b

2 = -9 +b

11 = b

y = -3x + 11 (substitute the new slope and y-intercept into the equation) is perpendicular to y = 1/3x + 2.

I hope that this example helps you understand how to solve perpendicular lines in the future. 

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

If ax+by+c  is the equation of a line, to find the equation of tlene perpendicular to ax+by+c = 0 we just interchange the coefficients of x and y and put a minus sign to one of the two.

So the equation of the line perpendicular line to the line ax+by+c is of the form, bx-ay + k1 = 0 Or -bx+ay + k2 = 0.

This is because ax+by +c = 0 has a slope a/b.

So any any line  perpendicular to ax+by+c = 0, must have slope , m  which , with a/b gives a product -1. Or m*(a.b) = -1. So m = -b/a.

We’ve answered 319,864 questions. We can answer yours, too.

Ask a question