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# Writing the Equation of Perpendicular Lines

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Two lines are said to be perpendicular if the angle between them is 90º (90 degrees). There is a result on perpendicular line which states: Two lines are perpendicular if and only if the product of their slopes is -1.

i.e. If the slopes of two lines which are perpendicular are m1 and m2 then m1 x  m2=-1.

Example:

The slope of a line is 3/2. If a line is perpendicular to this line, what is the slope of this new line?

Solution:

By the condition for the slopes of perpendicular lines,

m1 x  m2=-1

So, the slope of new line = -2/3.

### Finding the Equation of Perpendicular Lines:

For finding the equation of a line perpendicular to ax+by+c=0 and passing through the point (x1,y1), there are 2 methods.

#### Method 1:

(1)    Finding the slope of ax+by+c=0 which can be obtained by the expression –a/b.

i.e. slope of ax+by+c=0 is a/b.

(2)  Finding the slope of the line perpendicular to it.

i.e. the slope of the perpendicular line, m = b/a.

(3)  Then the corresponding perpendicular line’s equation can be found using point slope form.

i.e., the required equation is,

y-y1=m(x-x1)

Example:

Find the equation of the line perpendicular to 3x+y+3=0 and passing through (-1,2).

Solution:

The slope of given line, m = -3/1=-3

The slope of the perpendicular line = 1/3.

(x1,y1) = (-1,2).

The equation of perpendicular line is,

y-y1=m(x-x1)

y-2 = 1/3 (x+1)

3y-6 = x+1

x-3y+7=0.

#### Method 2:

(1)    Take the equation of line perpendicular to ax+by+c=0 as bx-ay+k=0 where k is a constant.

(2)    The value of ‘k’ can be found b substituting the given point (x1,y1) in the equation bx – ay+k=0.

(3)    Substitute ‘k’ back into the equation bx-ay+k=0, which is the required equation.

Example:

Find the equation of the line perpendicular to 3x+y+3=0 and passing through (-1,2).

Solution:

The equation of line perpendicular to 3x+y+3=0 is,

x-3y+k=0.

Here x=-1 and y=2

-1-3(2)+k=0

-1-6+k=0

k-7=0

k=7.

So the required equation of the perpendicular line is,

x-3y+7=0

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