Straight Lines Part 2

Video 1

Video 2

Finding Equations of a Line

  1. Finding an equation with 2 points/coordinates only (Video 1 Explanation)


We can find the gradient of a line by examining the equation, when it is written in the form of y = mx + c. But what if, there is no equation, rather we have two points which lie on the line. Then using the 2 points we can find the equation a line that goes through this point.
Question: Find the equation of the straight line that goes through the points (1,5) and (6,9).
We need to find the equation of the line that goes through these two points.
First we work out the gradient, we do this by using the formula:
m = (y1 – y0)/(x1 – x0)
So when we have two coordinates, we can label one coordinate as (x0 ,y0) and the other as (x1,y1).
Why is there a 0 and 1 underneath the x and y. These are called subscripts. Lets say you want to use the letter x to represent a number, however you want to use the letter x to represent another number, you cannot because you have already used the letter x. So instead of writing x again, we can use subscripts to differentiate the two terms.
In our case, we have two coordinates, and the formula that we use to find the gradient uses notation with subscripts.
It is just notation, this one way that people get scared of maths, just because of the way things are written.
If we say that (x1,y1) = (6,9) and (x0 ,y0) = (1,5)
That means x1 = 6, y1 = 9, x0 = 1, y0 = 5
Lets now find m by substitution.
m = (9 - 5)/(6 - 1) = (4)/(5) = 4/5

so
now we have a part of our straight line equation, as we have the gradient of the line, we have
y = (4/5)x + c
How do we find the y-intercept?
Well we have two coordinates that the line passes through, that means either coordinate will satisfy the equation of the straight line when we substitute values of x and y with either coordinate.
Lets use the coordinate (6,9)
So we know that when we substitute x = 6 and y = 9, the equation of the line should be satisfied
y = (4/5)x+ c
If we substitute x = 6 and y = 9, into the above equation we can solve and find c (the y-intercept).
9 = (4/5)(6) + c
9 = 24/5 + c
Subtract 24/5 from both sides
9 - 24/5 = 24/5 - 24/5 + c
9/1 - 24/5 = c
45/5 - 24/5 = c
21/5 = c
So
if we substitute c with 21/5, we get the equation of the straight line
y = (4/5)x + 21/5
Lets check whether this line goes through the coordinate (1,5), by seeing if our straight line equation is satisfied when we substitute x with 1 and y with 7 into the equation.
Does
5 = (4/5)(1) + 21/5
5 = 4/5 + 21/5 = 25/5 = 5 YES.
So the equation of the line that goes through the points (1,5) and (6,9) is y = (4/5)x + 21/5

  1. Finding the equation with a gradient and a point (Video 2 Explanation)


Given a gradient and a point, we can find the equation of a straight line.
Question: Find the equation of the line that goes through (-2,3) and has a gradient equal to 4.
Given that the equation of the straight line will be in the form of y = mx + c, we know what m is.
m is equal 4 (THIS IS GIVEN IN THE QUESTION)
We can substitute m with 4.
So
y = 4x + c
Now we need to find the y-intercept.
The question tells us that we need to find the equation of a line that goes through the point (-2,3), that means when x = -2 and y = 3 in our equation of a straight line, the equation MUST BE SATISFIED.
So we can substitute x with -2 and y with 3 in our equation to solve for c as shown below.
3 = 4(-2) + c
3 = -8 + c
(Add 8 to both sides)
3 + 8 = -8 + c + 8
11 = c
Given now we know the value of c, we can substitute c with 11 and thus attain our equation as shown below.
y = 4x + 11
Thus the equation of the line that goes through (-2,3) with gradient 4 is y = 4x + 11