Lucan Community College
Junior Certificate Mathematics
Class Rousseau & Camus – Mr Duffy
EQUATION OF A LINE
STANDARD FORMULA

To be learned off by heart like all formulae in
this chapter
y  y1  m( x  x1 )

Where ( x , y )is a point on the line and m
1 1
stands for the slope of the line
BASICS
Examining the formula you can see that we
need two things
 1. ONE point on the line
 2. The slope of the line

SUBSTITUTION
y  y1  m( x  x1 )
THESE 4 PARTS NEVER CHANGE
WE ONLY EVER CHANGE
X1, Y1 AND M.
EXAMPLE 1
y  y1  m( x  x1 )
x1 , y1
y  y1  m( x  x1 )
y  1  3( x  4)
Y1
M
X1

Multiplying this out and tidying up we get
y  1  3x  12

Always write equations of lines in the form
y  mx  c
This gives us y  3x  12  1
 So,

y  3x  13
NOT GIVEN THE SLOPE........
If we’re not given the slope of the line we will
simply have to calculate it. This can be done in
one of 3 simple ways.
y2  y1
 1. Use the slope formula
x2  x1

We must be given two points on the line to be
able to do this!
EXAMPLE 2

Find the slope of the line through a(3,-4) and
b(1,2). Hence find the equation of the line ab.
x1  3
y 2  y1
x2  x1

2  4
1 3
6
2
3
So, slope (m) = -3
y1  4
x2  1
y2  2
Equation: use the slope just found and the very
same x1 , y1 coordinates as used when
calculating the slope
 So we get............

y  y1  m( x  x1 )
y  4  3( x  3)
y  4  3 x  9
y  3 x  9  4
y  3 x  5
NOT GIVEN THE SLOPE........
 2.
A line could be parallel to the one
we’re looking for the equation of.
Remember, if a line is parallel to another one
then we can say that they have the same slope.
 We can easily get the slope of another line if
given its equation.


Write the equation in the form
y  mx  c
It is essential that there is no number in front
of the y whatsoever.
 If there is, then we divide the whole equation by
the number that is in front of it

EXAMPLE 3
Find the equation of the line through the point
(1,-6) which is parallel to the line
3x  y  4  0

[page 69 Q10 (i)]
 We need to find the slope, so when we’re given
another line’s equation we must rewrite it in the
form already mentioned i.e.
y  mx  c
 3x  y  4  0
so, m=number in front of x
y  3x  4
 Slope = m = 3


Now, the slope is 3 and the point we’re given
on the line is (1,-6) so let’s use these facts to
continue as normal.
y  y1  m( x  x1 )
y  6  3( x  1)
y  6  3x  3
y  3x  3  6
y  3x  9
NOT GIVEN THE SLOPE........
 3.
We can be given a line which is
perpendicular to the line we’re trying to
work out.
When this happens we find the slope the same as
before
 But, we use the PERPENDICULAR fact already
learned by turning the slope upside down and
changing the sign. This will given us the
perpendicular slope.

EXAMPLE 4



Find the equation of the line through the line (-2,1)
and which is perpendicular to the line 3x  2 y  4  0
[page 69 Q10 (ii)]
Start by finding the slope of the given line. We must
rewrite it first.
3
We can see that our slope is 2 as this is the
number in front of x. 3x  2 y  4  0
2 y  3 x  4
 3x 4

2
2
3
m
2
y
But this line is perpendicular so we must play
about with it first before using this slope in our
equation.
 Original slope =  3

2

To get perpendicular slope we turn it upside
down and change the sign, giving us
2
3

So, our slope is 2 and our point is (-2,1)
3


Now we get
Multiply across by
3 to get rid of the
fraction
y  y1  m( x  x1 )
2
y  1  ( x  2)
3
3 y  3  2( x  2)
3y  3  2x  4
3y  2x  4  3
3y  2x  7
TO PROVE A POINT IS ON A LINE
Simply substitute the point into the given
equation.
 Use brackets every time you sub in x and y and
when you work it all out the answer should be
0!

EXAMPLE 5
Verify that the point (-3,1) is on the line
2 x  4 y  2  0 [page 69, Q13]
 x  3, y  1

2(3)  4(1)  2  0
642  0
22  0
QED
HOW TO SKETCH A LINE
We need TWO points on a line to be able to draw it.
 We must have the equation of the line before we
can start.
 Let x=0 in this equation and find the
corresponding y-value. This is POINT 1.
 Now, let y=0 and find the corresponding x-value.
This is POINT 2.
 Draw a coordinate diagram and now plot these two
points.

PAGE 71 – QUESTION 5
Draw a sketch of the line 2 x  y  6  0 . Hence find
the area of the triangle formed by the x-axis,
the y-axis and the line.
 Step 1: Let x=0 in the given equation.
2(0)  y  6  0

So, our first point is (0,6)
0 y60

y6


Step 2: Let y=0 in the given equation.
2 x  (0)  6  0
2x  6  0
2 x  6
x  3
And, this is our 2nd point (-3,0)
NOW SKETCH THESE TWO POINTS USING
GRAPH PAPER
AREA OF THE TRIANGLE
Do you remember the formula
for the area of a triangle?
1
Base  Height
2
1
3 6
2
1
18
2
9units2
POINT OF INTERSECTION OF 2 LINES
To find the point of intersection of 2 lines we
use simultaneous equations. We cannot graph
these and read them from that. WE MUST use
Simultaneous Equations.
 Page 72 – Question 10
 Use simultaneous equations to find the point of
intersection of the lines

2 x  3 y  12
and
3 x  2 y  13.







2x-3y=12 Multiply this equation by 3
3x-2y=13
Multiply this equation by 2
6x-9y=36
6x-4y=26
Change the signs of the bottom line
6x-9y=36
-6x+4y=-26
Do exactly what it says on the tin!
0-5y=10
Divide both sides by -5.