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Co-ordinate geometry

Equation of a line between two points

Khan academy video on point-slope and standard form

There are many different methods of specifying the requirements for determining the equation of a straight line. One option is to find the equation of a straight line, when two points are given.

Assume that the two points are ( x 1 ; y 1 ) and ( x 2 ; y 2 ) , and we know that the general form of the equation for a straight line is:

y = m x + c

So, to determine the equation of the line passing through our two points, we need to determine values for m (the gradient of the line) and c (the y -intercept of the line). The resulting equation is

y - y 1 = m ( x - x 1 )

where ( x 1 ; y 1 ) are the co-ordinates of either given point.

Finding the second equation for a straight line

This is an example of a set of simultaneous equations, because we can write:

y 1 = m x 1 + c y 2 = m x 2 + c

We now have two equations, with two unknowns, m and c .

y 2 - y 1 = m x 2 - m x 1 m = y 2 - y 1 x 2 - x 1 y 1 = m x 1 + c c = y 1 - m x 1

Now, to make things a bit easier to remember, substitute [link] into [link] :

y = m x + c = m x + ( y 1 - m x 1 ) y - y 1 = m ( x - x 1 )

If you are asked to calculate the equation of a line passing through two points, use:

m = y 2 - y 1 x 2 - x 1

to calculate m and then use:

y - y 1 = m ( x - x 1 )

to determine the equation.

For example, the equation of the straight line passing through ( - 1 ; 1 ) and ( 2 ; 2 ) is given by first calculating m

m = y 2 - y 1 x 2 - x 1 = 2 - 1 2 - ( - 1 ) = 1 3

and then substituting this value into

y - y 1 = m ( x - x 1 )

to obtain

y - y 1 = 1 3 ( x - x 1 ) .

Then substitute ( - 1 ; 1 ) to obtain

y - ( 1 ) = 1 3 ( x - ( - 1 ) ) y - 1 = 1 3 x + 1 3 y = 1 3 x + 1 3 + 1 y = 1 3 x + 4 3

So, y = 1 3 x + 4 3 passes through ( - 1 ; 1 ) and ( 2 ; 2 ) .

Find the equation of the straight line passing through ( - 3 ; 2 ) and ( 5 ; 8 ) .

  1. ( x 1 ; y 1 ) = ( - 3 ; 2 ) ( x 2 ; y 2 ) = ( 5 ; 8 )
  2. m = y 2 - y 1 x 2 - x 1 = 8 - 2 5 - ( - 3 ) = 6 5 + 3 = 6 8 = 3 4
  3. y - y 1 = m ( x - x 1 ) y - ( 2 ) = 3 4 ( x - ( - 3 ) ) y = 3 4 ( x + 3 ) + 2 = 3 4 x + 3 4 · 3 + 2 = 3 4 x + 9 4 + 8 4 = 3 4 x + 17 4
  4. The equation of the straight line that passes through ( - 3 ; 2 ) and ( 5 ; 8 ) is y = 3 4 x + 17 4 .

Equation of a line through one point and parallel or perpendicular to another line

Another method of determining the equation of a straight-line is to be given one point, ( x 1 ; y 1 ) , and to be told that the line is parallel or perpendicular to another line. If the equation of the unknown line is y = m x + c and the equation of the second line is y = m 0 x + c 0 , then we know the following:

If the lines are parallel, then m = m 0 If the lines are perpendicular, then m × m 0 = - 1

Once we have determined a value for m , we can then use the given point together with:

y - y 1 = m ( x - x 1 )

to determine the equation of the line.

For example, find the equation of the line that is parallel to y = 2 x - 1 and that passes through ( - 1 ; 1 ) .

First we determine m , the slope of the line we are trying to find. Since the line we are looking for is parallel to y = 2 x - 1 ,

m = 2

The equation is found by substituting m and ( - 1 ; 1 ) into:

y - y 1 = m ( x - x 1 ) y - 1 = 2 ( x - ( - 1 ) y - 1 = 2 ( x + 1 ) y - 1 = 2 x + 2 y = 2 x + 2 + 1 y = 2 x + 3
The equation of the line passing through ( - 1 ; 1 ) and parallel to y = 2 x - 1 is y = 2 x + 3 . It can be seen that the lines are parallel to each other. You can test this by using your ruler and measuring the perpendicular distance between the lines at different points.

Inclination of a line

(a) A line makes an angle θ with the x -axis. (b) The angle is dependent on the gradient. If the gradient of f is m f and the gradient of g is m g then m f > m g and θ f > θ g .

In [link] (a), we see that the line makes an angle θ with the x -axis. This angle is known as the inclination of the line and it is sometimes interesting to know what the value of θ is.

Firstly, we note that if the gradient changes, then the value of θ changes ( [link] (b)), so we suspect that the inclination of a line is related to the gradient. We know that the gradient is a ratio of a change in the y -direction to a change in the x -direction.

m = Δ y Δ x

But, in [link] (a) we see that

tan θ = Δ y Δ x m = tan θ

For example, to find the inclination of the line y = x , we know m = 1

tan θ = 1 θ = 45

Co-ordinate geometry

  1. Find the equations of the following lines
    1. through points ( - 1 ; 3 ) and ( 1 ; 4 )
    2. through points ( 7 ; - 3 ) and ( 0 ; 4 )
    3. parallel to y = 1 2 x + 3 passing through ( - 1 ; 3 )
    4. perpendicular to y = - 1 2 x + 3 passing through ( - 1 ; 2 )
    5. perpendicular to 2 y + x = 6 passing through the origin
  2. Find the inclination of the following lines
    1. y = 2 x - 3
    2. y = 1 3 x - 7
    3. 4 y = 3 x + 8
    4. y = - 2 3 x + 3 (Hint: if m is negative θ must be in the second quadrant)
    5. 3 y + x - 3 = 0
  3. Show that the line y = k for any constant k is parallel to the x-axis. (Hint: Show that the inclination of this line is 0 .)
  4. Show that the line x = k for any constant k is parallel to the y-axis. (Hint: Show that the inclination of this line is 90 .)

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Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
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