On the higher paper you may be asked questions about parallel and perpendicular lines on graphs.
Parallel and perpendicular lines
On a graph, parallel lines have the same gradient.
For example, y = 2x + 3 and y = 2x - 4 are parallel because they both have a gradient of 2.
Remember that perpendicular lines will always cross at right angles.
In this diagram, the lines y = 2x + 3 and y = -1/2 x -1 cross at right angles.
The gradients of these lines are 2 and -1/2.
The product of the gradients is 2 x -1/2= -1.
You can work out whether 2 lines are perpendicular by multiplying their gradients. The product of the gradient of perpendicular lines will always be -1.
If lines are perpendicular M1× M2 = − 1
In exams, you will often be asked to find the equation of a line that is perpendicular to a given line. To do this, you will need to work out the gradient of one line before finding the gradient and equation of the other.
Example.
Find the perpendicular line to 4y - 3x = 8 through the point (0, 2).
- Re-arrange the equation 4y - 3x = 8 in the form y = mx + c
- y = 3/4x +2
The gradient is 3/4
Now we need to work out the gradient of the 2nd line. Remember that when 2 lines are perpendicular the product of their gradients is -1. Let's call the gradient of the second line m.
- 3/4m = -1
- m = -4/3
In the question we are told that the line passes through the point (0, 2). This means that the line crosses the y axis at +2.
So the equation of the line that is perpendicular to 4y - 3x = 8 is y = - 4/3x + 2
- Question
Line A goes through the points (4, 9 ) and (1, 3). Find the perpendicular line through the point (2, 0).
- Answer
First, find the gradient of line A. We know that the line passes through the points (1, 3) and (4, 9). To calculate the gradient, find the difference in the y co-ordinates and the difference in the x co-ordinates, and then divide the y value by the x value.
9 - 3/4 - 1 = 2.
The gradient of line A is 2.
We know that the product of the gradients of perpendicular lines is -1. If we call the gradient of line B m, then:
- m × 2 = -1
- m = -1/2
A straight line always has the equation y= mx + c
m is always the gradient so we know this equatuion is
y = -1/2x + c
Line B passes through the point (2, 0). To find out the value of c in the equation
y = -1/2x + c
we substitute the values in the equation with x = 2 and y = 0.
- 0 = -1/2 × 2 + c
- 0 = -1 + c
- c = 1
The equation of line B is therefore y = -1/2x + 1
Now try a Test Bite
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