Download Parallel and Perpendicular Slopes

Parallel and Perpendicular
Lines
y  mx  b
From
Intermediate
Algebra
Slope-Intercept Form
•Useful for graphing since m is the slope and b is the y-intercept
y  y1  mx  x1 
Point-Slope Form
•Use this form when you know a point on the line and the slope
•Also can use this version if you have two points on the line
because you can first find the slope using the slope formula and
then use one of the points and the slope in this equation.
ax  by  c  0
General Form
•Commonly used to write linear equation problems or express answers
1
m
3
1
m
3
3
m    3
1
1
m
3
The slope is a number that tells "how
steep" the line is and in which direction.
So as you can see, parallel lines have
the same slopes so if you need the
slope of a line parallel to a given line,
simply find the slope of the given line
and the slope you want for a parallel
line will be the same.
Perpendicular lines have negative
reciprocal slopes so if you need the
slope of a line perpendicular to a given
line, simply find the slope of the given
line, take its reciprocal (flip it over) and
make it negative.
Let's look at a line and a point not on the line
Let's find the equation of a
line parallel to y = - x that
y=-x
passes through the point (2, 4)
(2, 4)
What is the slope of the
first line, y = - 1x ?
This is in slope intercept
form so y = mx + b which
means the slope is –1.
y  4y1  m
-1 x  x21 
Distribute and then solve for
y to leave in slope-intercept
form.
y  x  6
So we know the slope is –1
and it passes through (2, 4).
Having the point and the
slope, we can use the pointslope formula to find the
equation of the line
What if we wanted perpendicular instead of parallel?
Let's find the equation of a
line perpendicular to y = - x
that passes through the point
y=-x
(2, 4)
(2, 4)
The slope of the first
line is still –1.
The slope of a line perpendicular
is the negative reciporical so
take –1 and "flip" it over and
make it negative.
y  y41  m
1 x  x21 
Distribute and then solve for
y to leave in slope-intercept
form.
y  x2
1
1
 1 
 
 1 
So the slope of a
perpendicular line is 1 and it
passes through (2, 4).