The graph of a linear relationship will create a line. All linear functions can be written in either of these two common forms:
Gradient intercept form | General form |
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$y=mx+c$y=mx+c | $ax+by+c=0$ax+by+c=0 |
If we sketch a linear relationship on a plane, the straight line formed will be one of the following types:
Regardless of all different shapes all linear functions have some common characteristics.
They all have at least one intercept. Linear functions might have
The $x$x intercept occurs at the point where $y=0$y=0.
The $y$y intercept occurs at the point where $x=0$x=0.
The gradient (slope) of a line is a measure of how steep the line is. For a linear function the gradient is constant. That is as the $x$x-value increases by a constant amount, the $y$y-value also increases by a constant amount. We can calculate the gradient from any two points $\left(x_1,y_1\right)$(x1,y1), $\left(x_2,y_2\right)$(x2,y2) on a line:
$m$m | $=$= | $\frac{rise}{run}$riserun |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |
The gradient is often represented by the letter $m$m and has the following properties:
On horizontal lines, the $y$y value is always the same for every point on the line. In other words, there is no rise- it's completely flat.
$A=\left(-4,4\right)$A=(−4,4)
$B=\left(2,4\right)$B=(2,4)
$C=\left(4,4\right)$C=(4,4)
All the $y$y-coordinates are the same. Every point on the line has a $y$y value equal to $4$4, regardless of the $x$x-value.
The equation of this line is $y=4$y=4.
Since gradient is calculated by $\frac{\text{rise }}{\text{run }}$rise run and there is no rise (ie. $\text{rise }=0$rise =0), the gradient of a horizontal line is always zero.
On vertical lines, the $x$x value is always the same for every point on the line.
Let's look at the coordinates for A,B and C on this line.
$A=\left(5,-4\right)$A=(5,−4)
$B=\left(5,-2\right)$B=(5,−2)
$C=\left(5,4\right)$C=(5,4)
All the $x$x-coordinates are the same, $x=5$x=5, regardless of the $y$y value.
The equation of this line is $x=5$x=5.
Vertical lines have no "run" (ie. $\text{run }=0$run =0). l If we substituted this into the $\frac{\text{rise }}{\text{run }}$rise run equation, we'd have a $0$0 as the denominator of the fraction. However, fractions with a denominator of $0$0 are undefined.
So, the gradient of vertical lines is always undefined.
A line has the following equation: $y=6\left(3x-2\right)$y=6(3x−2)
Rewrite $y=6\left(3x-2\right)$y=6(3x−2) in the form $y=mx+c$y=mx+c.
State the gradient and $y$y-value of the $y$y-intercept of the equation.
Gradient | $\editable{}$ |
Value at $y$y-intercept | $\editable{}$ |
Examine the graph attached and answer the following questions.
What is the slope of the line?
$0$0
$1$1
undefined
Does this line have a $y$y-intercept?
Yes
No
What is the $x$x-intercept of the line?
What is the gradient of the line going through A $\left(-1,1\right)$(−1,1) and B $\left(5,2\right)$(5,2)?
A line passing through the points $\left(5,3\right)$(5,3) and $\left(2,t\right)$(2,t) has a gradient equal to $-4$−4.
Find the value of $t$t.
We can think of the line as an infinite collection of points, all of which lie in a straight line and obey a given rule. The critical thing to remember is that the coordinates of every single point that is on the line, when substituted into the lines equation, will make that equation true. In other words, points on the line will satisfy the equation. Points not on the line will not satisfy the equation.
We can check if a point lies on a line by directly substituting the coordinates of the point into the equation of a line. If both sides are equal, then we claim that the point lies on the line. If both sides are not equal, then we claim that the point does not lie on the line.
Which of the following points lie on the line $y=9+\frac{x}{2}$y=9+x2?
$\left(4,11\right)$(4,11)
$\left(2,9\right)$(2,9)
$\left(11,4\right)$(11,4)
$\left(2,10\right)$(2,10)
To graph any linear relationship you only need two points that are on the line. We have three methods of finding those two points.
x | 1 | 2 | 3 | 4 |
---|---|---|---|---|
y | 3 | 5 | 7 | 9 |
To sketch from a table of values, we need just any two points from the table. From this table we have 4 coordinates, $\left(1,3\right)$(1,3), $\left(2,5\right)$(2,5), $\left(3,7\right)$(3,7), $\left(4,9\right)$(4,9).
Drag the $2$2 of the points on this interactive to the correct positions and graph this linear relationship.
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If we are given the equation of a linear relationship, like $y=3x+5$y=3x+5, then to sketch it we need two points. We can pick any two points we like.
Start by picking any two $x$x-values you like, often the $x$x-value of $0$0 is a good one to pick because the calculation for y can be quite simple. For our example, $y=3x+5$y=3x+5 becomes $y=0+5$y=0+5, $y=5$y=5. This gives us the point $\left(0,5\right)$(0,5)
Similarly look for other easy values to calculate such as $1$1, $10$10, $2$2. I'll pick $x=1$x=1. Then for $y=3x+5$y=3x+5, we have $y=3\times1+5$y=3×1+5, $y=8$y=8.This gives us the point $\left(1,8\right)$(1,8)
Now we plot the two points and create a line.
The general form of a line is great for identifying both the x and y intercepts easily.
For example, the line $3y+2x-6=0$3y+2x−6=0
The $x$x-intercept happens when the $y$y value is $0$0.
$3y+2x-6$3y+2x−6 | $=$= | $0$0 |
$0+2x-6$0+2x−6 | $=$= | $0$0 |
$2x$2x | $=$= | $6$6 |
$x$x | $=$= | $3$3 |
The $y$y-intercept happens when the $x$x value is $0$0.
$3y+2x-6$3y+2x−6 | $=$= | $0$0 |
$3y+0-6$3y+0−6 | $=$= | $0$0 |
$3y$3y | $=$= | $6$6 |
$y$y | $=$= | $2$2 |
From here it is pretty easy to sketch, we find the $x$x intercept $3$3, and the $y$y intercept $2$2, and draw the line through both.
Start by plotting the single point that you are given.
Remembering that gradient is a measure of change in the rise per change in run, we can step out one measure of the gradient from the original point given.
A line with gradient $4$4. Move $1$1 unit across and $4$4 units up. | A line with gradient $-3$−3. Move $1$1 unit across and $3$3 units down. | A line with gradient $\frac{1}{2}$12. Move $2$2 units across and $1$1 unit up. |
The point can be any point $\left(x,y\right)$(x,y), or it could be an intercept. Either way, plot the point, step out the gradient and draw your line!
For example, plot the line with gradient $-2$−2 and has $y$y intercept of $4$4.
First, plot the $y$y-intercept that has coordinates $\left(0,4\right)$(0,4) |
Step out the gradient, (-$2$2 means $1$1 unit across, $2$2 units down) |
Connect the points and draw the line.
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Consider the equation $y=-x+1$y=−x+1.
Complete the table of values for this equation:
$x$x | $-1$−1 | $0$0 | $1$1 | $2$2 |
$y$y | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Draw the line on the graph.
Determine whether the point $\left(1.5,-0.5\right)$(1.5,−0.5) lies on the line.
No
Yes
Consider the line with equation: $3x+y+2=0$3x+y+2=0
Solve for the $x$x-value of the $x$x-intercept of the line.
Solve for the $y$y-value of the $y$y-intercept of the line.
Plot the line.
Consider the linear equation $y=3x+1$y=3x+1.
State the $y$y-value of the $y$y-intercept of this line.
Using the point $Y$Y as the $y$y-intercept, sketch a graph of the equation $y=3x+1$y=3x+1.
Plot the line $x=-8$x=−8 on the number plane.
Consider the following linear equations:
$y=2x-4$y=2x−4 and $y=-2x-4$y=−2x−4
What are the gradient and $y$y-intercept of the line $y=2x-4$y=2x−4?
gradient | $\editable{}$ |
$y$y-value of $y$y-intercept | $\editable{}$ |
What are the intercepts of the line $y=-2x-4$y=−2x−4?
$x$x-value of $x$x-intercept | $\editable{}$ |
$y$y-value of $y$y-intercept | $\editable{}$ |
Plot the lines of the 2 equations on the same graph.
State the values of $x$x and $y$y which satisfy both equations.
$x$x = $\editable{}$
$y$y = $\editable{}$