#### Transcript Parallel and Perpendicular

```Parallel and Perpendicular
Lines
y  mx  b
Gradient-Intercept Form
•Useful for graphing since m is the gradient and b is the yintercept
y  y  m x  x Point-Gradient Form
1

1

•Use this form when you know a point on the line and the gradient
•Also can use this version if you have two points on the line because
you can first find the gradient using the gradient formula and then
use one of the points and the gradient 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 gradient is a number that tells "how
steep" the line is and in which direction.
So as you can see, parallel lines have
the same gradients so if you need the
gradient of a line parallel to a given
line, simply find the gradient of the given
line and the gradient you want for a
parallel line will be the same.
Perpendicular lines have negative
reciprocal gradients so if you need the
gradient of a line perpendicular to a
given line, simply find the gradient 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)
y  4y1  m
-1 x  x21 
Distribute and then solve for y
to leave in gradient-intercept
form.
y  x  6
What is the gradient of
the first line, y = -1x ?
This is in gradient
intercept form so
y = mx + b which means
the gradient is –1.
So we know the gradient is
–1 and it passes through
(2, 4). Having the point and
the gradient, we can use the
point-gradient 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 gradient of the first
line is still –1.
The gradient of a line
perpendicular is the negative
reciprocal so take –1 and "flip" it
over and make it negative.
y  y4 1  m
1 x  x21 
1
1
 1 
 
 1 
Distribute and then solve for y So the gradient of a
to leave in gradient-intercept
perpendicular line is 1 and it
form.
passes through (2, 4).
y  x2
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au
```