Parallel and perpendicular lines - Higher
On the higher paper you may be asked questions about parallel and perpendicular lines on graphs.
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.
Find the perpendicular line to 4y - 3x = 8 through the point (0, 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.
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
To find the gradient of a line we need to know how many it goes up, for every one across.
Find the gradient of the line joining (1,3) to (4,9).
As we go from (1,3) to (4,9) the y value increases by 6, and the x value increases by 3. So the line goes 6 up for 3 across. So this line has a gradient of 6/3 = 2.
Use this technique to answer the following question:
Line A goes through the points (4, 9) and (1, 3). Find the perpendicular line through the point (2, 0).
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 coordinates and the difference in the x coordinates, and then divide the y value by the x value.
9 - 3/4 - 1 = 6/3 = 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:
A straight line always has the equation y= mx + c
m is always the gradient so we know this equation 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.
The equation of line B is therefore y = -1/2x + 1