how to find a perpendicular line
Straight line graphs
y = mx + c is an important real-life equation. The gradient, m, represents rate of change (eg, cost per concert ticket) and the y-intercept, c, represents a starting value (eg, an admin. fee).
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 perpendicular if they meet at a right angle .
Two lines will be perpendicular if the product 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 reciprocal 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
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Find the equation of the line that is perpendicular to \(y = 3x - 1\) and goes through point (2, 5).
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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:
\[y = -\frac{1}{3}x + 5 \frac{2}{3}\]
how to find a perpendicular line
Source: https://www.bbc.co.uk/bitesize/guides/z9387p3/revision/4
Posted by: johnsgorry1949.blogspot.com
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