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.
Gradient
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?
$4$4
$0$0
Undefined
What is the $y$y-value of the $y$y-intercept of the line?
Does this line have an $x$x-intercept?
No
Yes
Given any two points on a line $\left(x_1,y_1\right)$(x1,y1), $\left(x_2,y_2\right)$(x2,y2) we have seen that we can calculate the gradient $m$musing:
$m$m | $=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |
If we took one of those points and any other point on the line with the coordinates$(x,y)$(x,y) we can set up the following equation:
$m$m | $=$= | $\frac{y-y_1}{x-x_1}$y−y1x−x1 |
We can rearrange this to produce a convenient way for finding the equation of a line given the gradient and a point $\left(x_1,y_1\right)$(x1,y1). This is called the point-gradient formula and is stated below.
Given a point on the line $\left(x_1,y_1\right)$(x1,y1) and the gradient $m$m, the equation of the line is:
$y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1)
Find the equation of the line passing through the two points $\left(-3,7\right)$(−3,7) and $\left(5,9\right)$(5,9).
First we need to determine the gradient as follows:
$m$m | $=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |
$m$m | $=$= | $\frac{9-7}{5--3}$9−75−−3 |
$m$m | $=$= | $\frac{2}{8}$28 |
$m$m | $=$= | $\frac{1}{4}$14 |
We can use either of the points given but using the point $\left(5,9\right)$(5,9) means we will only have positive integers to deal with.
$y-y_1$y−y1 | $=$= | $m\left(x-x_1\right)$m(x−x1) |
The point-gradient formula |
$y-9$y−9 | $=$= | $\frac{1}{4}(x-5)$14(x−5) |
Substituting the point and gradient |
$y-9$y−9 | $=$= | $\frac{1}{4}x-\frac{5}{4}$14x−54 |
Expanding the brackets |
$y$y | $=$= | $\frac{1}{4}x+\frac{31}{4}$14x+314 |
Writing the equation gradient-intercept form |
Now we have found the equation of the line, which we could also write as $x-4y+7=0$x−4y+7=0 in general form.
A line goes through the points $\left(0,-3\right)$(0,−3) and $\left(1,4\right)$(1,4).
What is the gradient of the line?
Write the equation of the line in the form $y=mx+c$y=mx+c.
A line passes through the points $\left(4,-6\right)$(4,−6) and $\left(6,-9\right)$(6,−9).
Find the gradient of the line.
Find the equation of the line by substituting the gradient and one point into $y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1).
To graph any linear relationship you only need two points that are on the line. You can substitute in any two values of $x$x into the equation and solve for corresponding $y$y-value to create your own two points. Often, using the intercepts is one of the easiest ways to sketch the line.
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 $-6x+2y-12=0$−6x+2y−12=0.
Find the $y$y-value of the $y$y-intercept of the line.
Find the $x$x-value of the $x$x-intercept of the line.
Sketch a graph of the line below.
Graph the linear equation $y=2x-2$y=2x−2 by determining any two points on the line.
Two lines that are parallel have the same gradient.
Thus two lines given by $y=m_1x+b$y=m1x+b and $y=m_2x+c$y=m2x+c will be parallel if and only if $m_1=m_2$m1=m2.
We can explain this feature with the following diagram:
Parallel property $m_1=m_2=\frac{p}{q}$m1=m2=pq |
The parallel property is self-evident. If the lines $PQ$PQ and $P'Q'$P′Q′ are parallel, then for a given run $q$q, the rise on both lines are equal. That is to say for both lines the gradient $m=\frac{p}{q}$m=pq.
So for example two lines parallel to $y=2x+3$y=2x+3 are $y=2x+7$y=2x+7 and $y=2x-12$y=2x−12.
Two lines are perpendicular if the product of their respective gradients equals $-1$−1. Another way to say this is that the gradient of one line is the negative reciprocal of the other.
So, two lines given by $y=m_1x+b$y=m1x+b and $y=m_2x+c$y=m2x+c will be perpendicular if and only if $m_1=-\frac{1}{m_2}$m1=−1m2.
Perpendicular property $m_1=\frac{-b}{a}$m1=−ba and $m_2=\frac{a}{b}$m2=ab |
For the perpendicular property, let the positive gradient of the line $BA$BA be $m_1=\frac{a}{b}$m1=ab. Now, imagine we rotate $BA$BA $90^\circ$90° anticlockwise so that the new line $BA'$BA′ is perpendicular to $BA$BA. From the symmetry in the diagram we see that the gradient of $BA'$BA′ is negative and given by $m_2=-\frac{b}{a}$m2=−ba.
Two lines perpendicular to $y=2x+3$y=2x+3 are $y=-\frac{1}{2}x+9$y=−12x+9 and $y=-\frac{1}{2}x-1$y=−12x−1.
If the line formed by equation $1$1 is parallel to the line formed by equation $2$2, fill in the missing value below.
Equation $1$1: $0=-3y-5x+6$0=−3y−5x+6
Equation $2$2: $y$y$=$= $\editable{}$ $x$x$-$−$1$1
Consider the line $5x-3y-9=0$5x−3y−9=0.
Find the $y$y-intercept of the line.
Find the equation of the line that is perpendicular to the given line and has the same $y$y-intercept. Express in general form.