C1: Parallel and Perpendicular
Lines
Learning Objective : to be able to
calculate the gradients of two lines and
identify the conditions for them to be
parallel or perpendicular
Starter:
1. Calculate the length of the line joining (5,
6) and (10, 18) and find the mid- point.
2. The triangle ABC has vertices A(-5, 0),
B(-1, 3) and C(2, 7). Determine the type
of triangle it is.
Parallel lines
If two lines have the same gradient they are parallel.
Show that the lines 3y + 6x = 2 and y = –2x + 7 are parallel.
We can show this by rearranging the first equation so that
it is in the form y = mx + c.
3y + 6x = 2
3y = –6x + 2
–6x + 2
y=
3
y = –2x + 2/3
The gradient, m, is –2 for both lines and so they are parallel.
Exploring perpendicular lines
Perpendicular lines
If the gradients of two lines have a product
of –1 then they are perpendicular.
In general, if the gradient of a line is m, then the gradient of
1
the line perpendicular to it is m .
Find the gradient of the line perpendicular to the line
joining the points A(–2, 2) and B(4, –1).
The gradient of the line AB is -1/2
So the gradient of the line perpendicular to this is 2.
Because -½ x 2 = -1
Task 1 : Work out the gradient of the line
that is perpendicular to these gradients
1.
2.
3.
4.
5.
6.
7.
3
½
-5
¾
-2/5
0.1
-0.5
Task 2 : Work out whether these pairs of
lines are parallel, perpendicular or neither
1.
2.
3.
4.
5.
6.
7.
y = 4x + 2
y = 2/3 x – 1
y = 1/5 x + 9
y = 5x – 3
4x – 5y + 1 = 0
3x + 2y – 12 = 0
5x – y + 2 = 0
y = -1/4 x – 7
y = 2/3 x – 11
y = 5x + 9
5x – y + 4 = 0
8x – 10y -2 = 0
2x + 3y -6 = 0
2x + 10y – 4 = 0
Perpendicular lines
If the gradients of two lines have a product
of –1 then they are perpendicular.
In general, if the gradient of a line is m, then the gradient of
1
the line perpendicular to it is m .
Find the equation of the perpendicular bisector of the
line joining the points A(–2, 2) and B(4, –1).
The perpendicular bisector of the line AB has to pass through
the mid-point of AB.
Let’s call the mid-point of AB point M, so
  2 + 4 2 + (  1) 
1
,
=
(1,
)
M is the point 

2
2
2


Perpendicular lines
The gradient of the line joining A(–2, 2) and B(4, –1) is
mAB =
1  2
4  (2)
=
3
6
= 
1
2
The gradient of the perpendicular bisector of AB is therefore 2.
Using this and the fact that is passes through the point (1, 21 )
we can use y – y1 = m(x – x1) to write
y
1
2
= 2( x  1)
2 y  1 = 4( x  1)
2y 1= 4x  4
2y  4x + 3 = 0
So, the equation of the perpendicular bisector of the line
joining the points A(–2, 2) and B(4, –1) is 2y – 4x + 3 = 0.
Task 3
1.
Find an equation of the line that passes through the
point (6, -2) and is perpendicular to the line y = 3x + 5.
2.
Find an equation of the line that passes through the
point (-2, 7) and is parallel to the line y = 4x + 1. Write
your answer in the form ax + by + c = 0.
3.
The line r passes through the points (1, 4) and (6, 8)
and the line s passes through the points (5, -3) and
(20, 9). Show that the lines r and s are parallel.
4.
The vertices of a quadrilateral ABCD have coordinates
A(-1, 5), B(7, 1), C(5, -3) and D(-3, 1). Show that the
quadrilateral is a rectangle.