Parallel lines are lines that have the same gradient.
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 |
$ax+by+c=0$ax+by+c=0 | Parallel lines have the same value of $\frac{-a}{b}$−ab. |
$x+2y-3=0$x+2y−3=0 $2x+4y+1=0$2x+4y+1=0 |
For every straight line $y=mx+c$y=mx+c, there exist infinitely many lines parallel to it.
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.