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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.
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.
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