**What is a straight line?**

This seems like a silly question but in Mathematics, there are some misconceptions about what I straight line really is. A

**Straight Line**is any line that connects two points together but it is important to note any restrictions on the length of the line because it can either have a finite length (specific length) or it can extend infinitely and never end.

**The Equation of a Straight Line**

**$$y=mx+c$$**

This is the standard equation of a straight line where:

$$\text{m = the gradient / slope of the line}$$

$$\text{c = the y-intercept of the line} $$

$$\text{y = the y-coordinate of a specific}$$

$$\text{point on the graph}$$

$$\text{x = the corresponding x-coordinate}$$

$$\text{at the point on the graph}$$

**The Cartesian Plane**

I will give you two examples to show what I mean:

**Plotting points on the Cartesian Plane**

**If you are plotting a point on the Cartesian Plane, you would write the point as follows:**

$$(x,y)$$

where we replace x with the x-coordinate of the point and y with the y-coordinate of the point.

Example: Lets plot the points $$ A(0,0), B(3,6), C(5,-2), D(-3,4),$$ $$E(-2,-3) $$ You would sketch the Cartesian Plane and the points would be plotted as follows:

**Quadrants**

**There are 4 quadrants in the 2-dimensional Cartesian Plane. They are as follows:**

If you look at the points from the example above:

\( A(0,0) \) is in between all the quadrants so we generally just work with it as if it is in Quadrant 1 (it does not matter).

\( B(3,6) \) is in Quadrant 1.

\( C(5,-2) \) is in Quadrant 4.

\( D(-3,4) \) in in Quadrant 2.

\( E(-2,-3) \) is in Quadrant 3.

**Plotting a Straight Line on the Cartesian Plane**

In order to plot a straight line on the Cartesian Plane we need to plot at least 2 points and then draw a straight line through these points.

By convention, we try and find the

**x-intercept**(where the graph cuts the x-axis, \( y=0 \) ) and**y-intercept**(where the graph cuts the y-axis, \( x=0\) ) as these points.
Example: Find the equation of the line \( y = -3x + 9 \)

**y-intercept (x=0)**

$$y = -3(0) + 9$$

$$y=9$$

$$(0,9)$$

**x-intercept (y=0)**

$$0 = -3x+9$$

$$3x = 9$$

$$\frac{3x}{3} = \frac{9}{3}$$

$$x=3$$

$$(3,0)$$

So when we plot these points we get:

Now we draw a straight line through these 2 points and label it \( y = -3x + 9 \):

**Note that**if they give you only 1 point and either m or c then you can substitute the point and solve the equation to get m or c and then you will have the equation of the line and you can draw the graph the same way as above.

Example: \( E(3,4) \) is a point on the line \( y = -x + c \). Find c and then sketch the graph.

**Substitute the point E(3,4) into the equation**

$$4 = -(3) + c$$

$$4 = -3 + c$$

$$4+3 = c$$
$$c=7$$

\( y = -x + 7 \) is the equation of the graph

**y-intercept (x=0)**

$$y= -(0) + 7$$

$$y=7$$

$$(0,7)$$

**x-intercept (y=0)**

$$0= -x + 7$$

$$x=7$$

$$(7,0)$$

So the graph looks like:

**Finding the Equation of a Straight Line by using the Graph**

You need to have 2 points (or be able to see exactly where they are and could be intercepts but do not need to be) so that we can calculate the gradient. You may be able to see the y-intercept but we can also find it by substitution.

Example: You are given 2 points \( E(-1, -\frac{9}{7}) \) and \(F(2, \frac{33}{7}) \). Find the equation of the straight line that goes through these 2 points.

**Plot the points and draw a line through them**

**Calculate the gradient / slope**

The

**Gradient (m)**of a straight line is the slope characterized by the change in y-coordinates over the change in x-coordinates. It is always constant (the same) when dealing with a straight line!
$$m = \frac{\Delta y}{\Delta x}$$

where \( \Delta \) means change.

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

In this Example, we need to choose which point is the 1st and which is the 2nd (It doesn't matter). Say E is the 1st and F is the 2nd. We write:

$$m_{EF} = \frac{y_F - y_E}{x_F-x_E}$$

if you say \( m_{FE} = \frac{y_E - y_F}{x_E - x_F} \) you will get the same answer. Try it!

$$m_{EF} = \frac{\frac{33}{7} - (- \frac{9}{7})}{2 - (-1)}$$

$$= \frac{(\frac{42}{7})}{3}$$

$$\frac{6}{3} = 2$$

$$y = 2x + c$$

**Substitute in a point (any). Say \( E(-1, -\frac{9}{7}) \)**

$$-\frac{9}{7} = 2(-1) + c$$

$$-\frac{9}{7} = -2 + c$$

$$-\frac{9}{7} + 2 = c$$

$$c = -\frac{9}{7} + \frac{14}{7}$$

$$c = \frac{5}{7}$$

$$y = 2x + \frac{5}{7}$$

which is the equation of the straight line so we are finished.

**Restrictions on the Domain**

The way we have been drawing these straight lines uses an assumption that \( x \in \mathbb R \) (i.e. x is any element of the real number system).

Lets use a simple example of \( y = x + 1 \)

**Note that**x can be any real number and it extends to numbers you cannot see on the graph by showing with arrows that the graph continues.

**Restrictions on x**

Lets say for example:

a) x = {-1, 0 , 1} then the graph would look like:

In this case. These 3 points make up the entire graph with a restricted domain.

b) \( x \in [-1,1] \) or \( -1 \leq x \leq 1 \). The graph would look like:

where the graph does not continue.

**Note that**similar restrictions are possible on the

**y-values**but when we restrict the y-values, we say we are restricting the

**range**instead of domain.

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