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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