**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 a Straight-Line Equation?**

A 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)*

Source of image: https://thirdspacelearning.com/gcse-maths/algebra/y-mx-c/

**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.

y^{2} – y^{1}

________

x^{2} – x^{1}

** 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

**or**

**Eq 1****, and solve for c (the y-intercept). For example, using**

**Eq 2**

**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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