Finding the equation of a line II
So far you have learned that the equation of a line passing through a
point (x1 , y1 ) , with gradient m, is y y1 = m(x x1 ) . This is because
there is a unique line passing through any given point with a particular
gradient.
There is also only one line that can pass through any two given points.
Two-point formula
When you are given the coordinates of two points on a line you can find
the equation of that line.
A(x1 , y1 ) and B(x2 , y2 ) are two given
y
points with known coordinates.
Let P(x, y) be any other point on this
line.
P(x, y)
A(x1, y1)
Using the gradient formula you have
y y1
m=
x x1
m=
y2 y1
x2 x1
x
(consider points P and A)
B(x2, y2)
(consider points B and A)
A basic property of a straight line is that its gradient remains constant.
y – y1
y –y
= x2 – x1 is the equation of the line
x – x1
2
1
which passes through the two points
(x1, y1) and (x2, y2).
Part 1
Equations of a straight line
1
You can use this equation instead of needing to calculate the gradient
between two points first.
Follow through the steps in this example. Do your own working in the
margin if you wish.
Find the equation of the line that passes through the two points
P(1, –3) and Q(–2, 5).
Solution
y y1
x x1
=
y2 y1
x2 x1
(x1 , y1 )
(x2 , y2 )
P(1, 3) Q(2, 5)
y (3) 5 (3)
=
2 1
x 1
y+3 8
=
x 1 3
3( y + 3) = 8(x 1)
[multiply both sides by 3(x 1)]
3y + 9 = 8x + 8
8x + 3y +1 = 0
It is always a good idea to write down the equation
first before substituting values. That way,
• you learn the equation because you have
written it out a number of times
• you can see what you are substituting into,
and are less likely to make a mistake
• you show your teacher, and others, you know
which formula to use in a particular question.
2
PAS5.3.3 Coordinate geometry
Activity – Finding the equation of a line II
Try these.
1
Write the equation, in general form, of the line that passes through
the points (–2, 4) and (1, 5).
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2
A line cuts the x-axis at –3 and the y-axis at 5. Use the two point
form of the equation of a straight line to find its equation.
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Check your response by going to the suggested answers section.
In using the two-point form it does not matter which point you label as
point 1 and which point 2. The answer will still be the same.
Part 1
Equations of a straight line
3
Activity – Finding the equation of a line II
1
y y1
x x1
=
y2 y1
x2 x1
5 4
y4
=
x (2) 1 (2)
y4 1
=
x+2 3
3( y 4) = 1(x + 2)
3y 12 = x + 2
x 3y +14 = 0
2
Points are (–3, 0) and (0, 5)
y y1
x x1
=
y2 y1
x2 x1
50
y0
=
x (3) 0 (3)
5
y
=
x+3 3
3y = 5(x + 3)
3y = 5x +15
5x 3y +15 = 0
3
From a sketch, m = 2
and b = 4.
3
y = mx + b
2
y = x + 4, or
3
2x + 3y 12 = 0.
4
4
x y
+ =1
a b
y
x
+
=1
4 3
3x 4 y = 12
3x + 4 y +12 = 0
PAS5.3.3 Coordinate geometry