Parallel lines are lines that have the same gradient.
Two parallel lines that never cross and don't have any points in common.
Let's look at how we can identify parallel lines given their equations.
Equation Form | Characteristic of parallel lines | Examples |
---|---|---|
$y=mx+c$y=mx+c | Parallel lines have the same $m$m value. |
$y=2x-1$y=2x−1 $y=4+2x$y=4+2x |
For every straight line $y=mx+c$y=mx+c, there exist infinitely many lines parallel to it.
Here is the line $y=x$y=x
Here are two more lines in the same family of parallel lines: $y=x+3$y=x+3 and $y=x-6$y=x−6.
Notice that they have the same gradient ($m$m-value) but different $x$x and $y$y-intercepts ($c$c-values).
Horizontal lines are lines where the $y$y-value is always the same.
Let's look at the coordinates for $A$A, $B$B and $C$C on this line.
$A=\left(-8,4\right)$A=(−8,4)
$B=\left(-2,4\right)$B=(−2,4)
$C=\left(7,4\right)$C=(7,4)
All the $y$y-coordinates are the same, $y=4$y=4.
This means that regardless of the $x$x-value the $y$y value is always $4$4.
The equation of this line is $y=4$y=4
So if the equation of a straight line is $y=c$y=c, then it will be a horizontal line passing through the point $\left(0,c\right)$(0,c).
The $x$x-axis itself is a horizontal line. The equation of the $x$x-axis is $y=0$y=0.
All horizontal lines are parallel to the $x$x-axis and are of the form $y=c$y=c.
They have a gradient of $0$0.
Vertical lines are lines where the $x$x-value is always the same.
Let's look at the coordinates for $A$A, $B$B and $C$C on this line.
$A=\left(-3,8\right)$A=(−3,8)
$B=\left(-3,3\right)$B=(−3,3)
$C=\left(-3,-3\right)$C=(−3,−3)
All the $x$x-coordinates are the same, $x=-3$x=−3.
This means that regardless of the $y$y-value the $x$x-value is always $-3$−3.
The equation of this line is $x=-3$x=−3
So if an equation of a straight line is $x=c$x=c, then it will be a vertical line passing through the point $\left(c,0\right)$(c,0).
The $y$y-axis itself is a vertical line. The equation of the $y$y-axis is $x=0$x=0.
All vertical lines are parallel to the $y$y-axis and are of the form $x=c$x=c.
Their gradient is undefined.
Lines that meet at right angles ($90^\circ$90°) are called perpendicular lines.
Play with this applet creating pairs of perpendicular lines.
Fill in this table as you go.
Gradient of line 1 | $m_1$m1 | |||
---|---|---|---|---|
Gradient of line 2 | $m_2$m2 | |||
Product of line 1 and line 2 | $m_1\times m_2$m1×m2 |
What do you notice about the product of the gradients of lines $1$1 and $2$2? (The pair of perpendicular lines)
You will have discovered the perpendicular lines have gradients whose product is equal to $-1$−1.
We say that $m_1$m1 is the negative reciprocal of $m_2$m2.
Negative reciprocal is a complex sounding term, but it just means two numbers that have opposite signs and are reciprocals of each other.
Here are some examples of negative reciprocals:
$2$2 and $-\frac{1}{2}$−12
$\frac{3}{4}$34 and $-\frac{4}{3}$−43
$-10$−10 and $\frac{1}{10}$110
Which lines are parallel to $y=-3x+2$y=−3x+2?
Select the two correct options.
$y=3x$y=3x
$y=-\frac{2x}{3}+8$y=−2x3+8
$-3y-x=5$−3y−x=5
$y=-10-3x$y=−10−3x
$y+3x=7$y+3x=7
Write down the equation of a line that is parallel to the $x$x-axis and passes through $\left(-10,2\right)$(−10,2).
Consider the following points on the number plane:
$A$A $\left(2,-1\right)$(2,−1)
$B$B $\left(4,-7\right)$(4,−7)
$C$C $\left(-3,1\right)$(−3,1)
$D$D $\left(-6,10\right)$(−6,10)
First, calculate the gradient of the line $AB$AB.
Now, find the gradient of the line $CD$CD.
Is the line $CD$CD parallel to $AB$AB?
Yes
No
Assess whether the points $A$A, $B$B and $C$C are collinear.
If $A$A and $B$B have the coordinates $\left(-4,3\right)$(−4,3) and $\left(-2,7\right)$(−2,7) respectively, evaluate the gradient of $AB$AB.
If $C$C has the coordinates $\left(-7,-3\right)$(−7,−3), evaluate the gradient of $BC$BC.
Based on these two gradients, are $A$A, $B$B and $C$C collinear?
Yes
No
A line goes through A$\left(3,2\right)$(3,2) and B$\left(-2,4\right)$(−2,4):
Find the gradient of the given line.
Find the equation of the line that has a $y$y-intercept of $1$1 and is parallel to the line that goes through $A$A$\left(3,2\right)$(3,2) and $B$B$\left(-2,4\right)$(−2,4).
A line which passes through the point $\left(0,6\right)$(0,6) is perpendicular to $y=-3x+5$y=−3x+5.
Find the gradient of this perpendicular line.
State the equation of the perpendicular line.