Equation of a Straight Line

•  Edexcel - C1, Coordinate Geometry
• AQA - C1, Coordinate Geometry
• OCR -C1, Coordinate Geometry

Now we move on to Chapter 2, which involves Straight Lines in the x,y plane. As we know we can write linear equations in the form: y =mx + c, where m is the gradient and c is the y-intercept. We can write linear equations in two other forms one is ax + by + c = 0, where a,b,c are real numbers. Also we can rewrite it in the form y - y1 = m(x -x1), where x1,y1 are the x,y coordinates of a point on the line, and m being the gradient.


The latter form will be used a lot, as we need to be able to find the equation of a straight line, when we are given
a) Two points on the line
b) The Gradient and a point

To rearrange from the form y = mx + c, to ax + by + c = 0, is simple. Let's use an example :
If y = 3x + 5, where m =3 and c=5... Rearrange into the ax + by + c form :
3x - y + 5 =0, Done, we took y to the other side.
a = 3, b = -1 , and c=5.

We use this form because when y = mx + c, contains fractions it can look not so concise. This form is better in that respect.

Given two points

When we are given two point, and we need to find the equation of a line through the points. It is fairly simple.
1) First find the gradient of the line
Remember Gradient = dy/ dx (change in y / change in x)
This should give you the m, in the form y= mx +c.

2) Using one the points, use the equation form [  y - y1 = m(x-x1)  ], where x1, and y1 are the x,y coordinates of the point, and m is the gradient.
3) Substitute in the values, and multiply out the brackets, and simplify either in the form y=mx + c OR ax + by + c=0, usually stated in the question.

Example

Find the equation of a straight line through the points (3,7) and (4,10)

1) Find Gradient = (10 -7) / (4- 3)

= 3 /1

=3

m is 3.

2) Use of the points (3,7) or (4,10) and the gradient (3), substitute into the formula : y-y1 = m(x-x1)

Let's use 3,7

y-7 = 3(x-3)  <<Substitute in values

y-7 = 3x - 9   << Expand out

y = 3x - 9 + 7 << Simplify and collect like terms

y = 3x -2  << Equation in the form y = mx + c (m = 3, c = -2)

3x - y - 2 = 0  << Equation in the form ax + by + c   (a=3,  b= -1,  c = -2)

Given the Gradient and a Point

If a given the gradient, and a point its much easier. We just repeat the process using the formula, infact it's the same thing above, but easier, as we don't have to find the gradient. So just plug in the gradient and the coordinates in the formula : y - y1 = m(x- x1)

Perpendicular Lines and Parellel Lines 

* When two lines are parellel, the gradient of the lines is the same. (m)
* When two lines are perpendicular (at right angles), the product of the two gradients is equal to -1 .

Example

a) y = 4x - 3, find the equation of a line which is parellel :
Answer = Any line which has the gradient 4... y=4x + ... the y intercept does not matter.
b) Find the gradient of the line perpendicular :
Answer = Remember the product of the gradients has to be equal to -1.
So Let m be the gradient of the perpendicular line :
m x 4 = -1
m = -1/4