Coordinate Geometry Lesson #3
Equations of Lines
• The equation of a line is an equation that connects the x and y
values for every point on the line.
• Recall that: y = mx + c is the equation of a line with gradient m
and y-intercept c.
• Example:
1
y − step
and the
=
2
x − step
1
y-intercept is 1. Therefore, its equation is y = x + 1.
2
The illustrated line has gradient =
Finding The Equation Of A Line
• Consider the line below that has gradient
1
and passes
2
through the point (2, 3).
Suppose (x, y) is any point on the line. The gradient between
y−3
.
(2, 3) and (x, y) is
x−2
y−3
1
=
{gradient formula}
x−2
2
Consider two rearrangements of this equation of the line.
Equating gradients gives us
y−3
1
=
x−2
2
1
y – 3 = (x – 2) {multiplying both sides by x – 2}
2
1
y – 3 = x – 1 {expanding the bracket}
2
1
y= x+2
{adding 3 to both sides}
2
and this is in the form y = mx + c called the
gradient-intercept form.
OR
y−3
1
=
x−2
2
2⎛ y − 3 ⎞ 1⎛ x − 2⎞
=
2 ⎜⎝ x − 2 ⎟⎠ 2 ⎜⎝ x − 2 ⎠⎟
2(y – 3) = 1(x – 2)
2y – 6 = x – 2
x – 2y = -4
{as LCD is 2(x – 2)}
{equating numerators}
{expanding brackets}
{rearranging}
and this is in the form Ax + By = C, called the general form. In
this case, A = 1, B = -2, and C = -4.
• So, to find the equation of a line we need to:
o know (or be able to find) the gradient and
o the coordinates of any point on the line.
• Summary:
If a straight line has gradient m and passes through (a, b) then
y −b
=m
it has equation
x−a
which can be rearranged into:
y = mx + c
{gradient-intercept form}
or
Ax + By = C
{general form}.
• Example:
Find, in gradient-intercept form, the equation of the line through
(-1, 3) with a gradient of 5.
The equation of the line is:
y−3
=5
x − ( −1)
y−3
=5
x +1
y – 3 = 5(x + 1)
y – 3 = 5x + 5
y = 5x + 8
Therefore, the equation of the line is y = 5x + 8.
Gradient From The Equation Of The Line
1
2
x + , and
3
3
y = 5 – 2x, we can easily find the gradient by looking at the
coefficient of x. But, how do we find the gradients of equations
of lines in the general form? One method is to rearrange them.
• Example:
Find the gradient of the line 2x + 5y = 17.
• From the equations of lines such as y = 2x – 3, y =
2x + 5y = 17
5y = 17 – 2x
17 2x
−
y=
5
5
2
17
y= − x+
5
5
{subtracting 2x, both sides}
{dividing both sides by 5}
2
Therefore, the gradient is − .
5
Does A Point Lie On A Line?
• A point lies on a line if its coordinates satisfy the equation of the
line.
• Example:
Does (3, -2) lie on the same line with equation 5x – 2y = 20?
Substituting (3, -2) into 5x – 2y = 20 gives
5(3) – 2(-2) = 20
19 = 20
which is false.
Therefore, (3, -2) does not lie on the line.
Equations From Graphs
• Provided that a graph contains sufficient information, we can
determine its equation. Remember that we must have at least
one point and we must be able to determine its gradient.
• Example 1:
Find the equation of the line with the graph:
(a.)
(b.)
(a.)
Two points on the line are (0, 2) and (4, 3).
3−2
1
=
gradient m =
4
4−0
and the y-intercept, c = 2.
Therefore, the equation is y =
and y –
1
x + 2 in gradient-intercept form
4
1
x = 2 in general form.
4
(b.)
Two points on the line are (1, 2) and (5, 0).
0−2
−2
1
=
= −
gradient m =
4
2
5 −1
As we do not know the y-intercept, we use the following:
y−2
1
= −
equation is
x −1
2
2(y – 2) = -1(x – 1)
2y – 4 = -x + 1
x + 2y = 5
Therefore, the equation is x + 2y = 5 in general form and
1
y = − x + 5 in gradient-intercept form.
2
• Example 2:
Find the equation connecting the variables in:
(0, 3) and (5, 2) lie on the straight line.
2−3
1
gradient m =
= −
5−0
5
and the y-intercept, c = 3.
The equation is of the form Y = mX + c, where Y = N, X = p.
1
Therefore, the equation is N = − p + 3.
5