Parallel and perpendicular lines
Parallel lines
Parallel lines are a fixed distance apart and will never meet, no matter how long they are extended. Lines that are parallel have the same gradient.
The graphs above, \(y = 2x + 1\) and \(y = 2x - 2\) have the same gradient of 2.
The lines are parallel.
Example
State the equation of a line that is parallel to \(y = 3x + 7\).
To be parallel, two lines must have the same gradient. The gradient of \(y = 3x + 7\) is 3.
Any line with a gradient of 3 will be parallel to \(y = 3x + 7\).
Two examples are \(y = 3x - 2\) and \(y = 3x + 11.6\).
Perpendicular graphs - Higher
Two lines are If the angle between two lines is a right angle, the lines are said to be perpendicular. if they meet at a right angle.
Two lines will be perpendicular if the The answer when two or more numbers are multiplied together, eg the product of 4 and 15 is 60. of their gradients is -1.
To find the equation of a perpendicular line, first find the gradient of the line and use this to find the equation.
Example
Find the equation of a straight line that is perpendicular to \(y = 2x + 1\).
The gradient of \(y = 2x + 1\) is 2.
To find the perpendicular gradient, find the number which will multiply by 2 to give -1. This is the negative The reciprocal of a number is 1 divided by that number. The reciprocal of a fraction is that fraction turned upside down, eg the reciprocal of 3/4 is 4/3. of the gradient.
The reciprocal of 2 is \(\frac{1}{2}\), so the negative reciprocal of 2 is \(-\frac{1}{2}\).
This gives \(y = - \frac{1}{2}x + c\).
Examples of equations of lines that are perpendicular to \(y = 2x + 1\) would include \(y = -\frac{1}{2}x + 5\) or \(y = -\frac{1}{2}x - 4\).
Question
Find the equation of the line that is perpendicular to \(y = 3x - 1\) and goes through point (2, 5).
First, find the perpendicular gradient and substitute this into the equation for all straight lines, \(y = mx + c\).
If \(y = 3x - 1\) the gradient of the graph is 3. This means the gradient of the perpendicular graph is \(-\frac{1}{3}\) (by finding the negative reciprocal).
\(y = -\frac{1}{3}x + c\)
To find the value of \(c\), the question states that the graph goes through the point (2, 5). This means \(x = 2\) and \(y = 5\). Substitute these values into \(y = mx + c\).
\(y = -\frac{1}{3}x + c\)
\(x = 2\) and \(y = 5\).
\(5 = -\frac{1}{3} \times 2 + c\)
\(5 = -\frac{2}{3} + c\). Add \(\frac{2}{3}\) to each side to find \(c\)
\(5 \frac{2}{3} = c\)
The final equation is:
\(y = -\frac{1}{3}x + 5 \frac{2}{3}\)
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