In the national 5 maths course, students are expected to be able to solve straight line equations. There are two equations that can be used to solve straight-line equations, which are shown below. Normally the equation of a straight line is y = mx + c.
What is the Straight-Line Equation?
The straight-line equation can be used to work out the gradient and intercept of a linear equation. The general equation of the straight line is y = mx + c, where m = gradient and c is the y-intercept. The gradient can be calculated using any 2 points selected on the line of the graph. There is another form of the straight-line equation, which is y-b = m(x-a). Both equations will give the same answer, there is just a different approach taken for each to solve the problem. Equations 1 & 2 show the different equations used.
Eq 1: y = mx + c Eq2: y – b = m(x – a)
How to work out the gradient of a straight line
The gradient is the slope of the straight line. It can be calculated when there are two known points of the line, stating the x and y coordinates. Using the two points on the line, the following equation can be used to calculate the gradient (Eq 3). Where y2 and y1 are the y-coordinates, and x1 and x2 are the x-coordinates of the points on the line.
y2 – y1
x2 – x1
Eq3: Calculation of gradient
Example: A line passes through the points (1,3) and (3,7). Find the equation of the line.
Step 1: Find the gradient of the line by using the two points given.
Eq 3: m = 7 – 3/3 – 1 m = 4/2 = 2
Using Eq 3, the gradient can be determined, which is the difference in y-coordinates divided by the difference x-coordinates. Now that we have the gradient, we can move on to step 2 and work out the y-intercept of the equation.
Step 2: Substitute one of the points on the line, into either Eq 1 or Eq 2, and solve for c (the y-intercept). For example, using Eq 1 (y = mx + c), a point (x,y) on the line can be substituted into the equation, which can then be used to solve for the value of c (the intercept).
Sub in point (1,3) into y = mx + c:
y = mx + c
3 = (2×1) + c
3 = 2 + c
c = 1
y = 2x +1
Sub in point (1,3) into y-b = m(x-a):
y – b = m(x – a)
y – 3 = 2(x – 1)
y – 3 = 2x – 2
y = 2x + 1
As you can see from the example above both equations give you the same final answer. Follow the video link below to see a full explanation of how to solve the straight-line equation using the two different straight-line equations.
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