We have learned the concept of lines. When considering two lines, we get a particular form of lines. Like the type of lines, you get to see on the railway track crossing sign, intersecting edges of floor and wall, or the plus sign on the first aid kit. These types of lines are perpendicular lines.
Here we will take a look at perpendicular lines and understand the different concepts related to them.
Perpendicular lines meaning
Perpendicular lines are the lines that intersect each other at a certain angle. As the name says, a perpendicular is formed between the two lines. Perpendicular is a right angle. Hence, both lines intersect at \(90º\).
Two distinct straight lines which intersect at \(90º\) are called perpendicular lines.
Perpendicular lines, StudySmarter Originals
Here straight lines AB and CD intersect at point O and that intersecting angle is \(90\) degrees. So both the lines \(AB\) and \(CD\) are perpendicular lines. So, we denote them with a sign \(\perp\).
\[\implies AB\perp CD\]
Also, remember that all the four angles in perpendicular lines will be equal to \(90\) degrees. So, here
\[\angle AOD=\angle AOC=\angle COB=\angle BOD=90º\]
Non-perpendicular lines, StudySmarter Originals
Here above both types of lines are not perpendicular lines as the lines in the first figure intersect but not at \(90º\). And the lines in the second figure do not intersect at all. Therefore, one should note that not all intersecting lines are perpendicular lines.
Perpendicular lines Gradient
The gradient of perpendicular lines is the slope or the steepness of the lines. As both the perpendicular lines are, in fact, a line in itself, we can represent them in the form of a line equation \(y=mx+b\). This equation describes the value of \(y\) as it varies with \(x\). And m is the slope of that line and \(b\) is the y-intercept.
The slope of the perpendicular lines is the negative reciprocal of each other. Suppose the slope of the first line is \(m_1\) and the slope of the second line is \(m_2\). The relation between both the perpendicular line slope is \(m_1 ·m_2=-1\).
Hence, we can say that if the product of two slopes is \(-1\) then both the lines are perpendicular to each other.
Perpendicular lines with gradient relation, StudySmarter Originals
Perpendicular line slope formula
We can find the slope of the perpendicular line with the help of the equation of a line and using the above-mentioned concept of slope. The general form of the equation of a line is represented as \(ax+by+c=0\). Then we can simplify this equation as:
\[ax+by+c=0\]
\[\implies y=-\dfrac{a}{b}x-\dfrac{c}{b}\quad \quad (1)\]
We also know that the equation of a line in terms of slope can be written as,
\[y=m_1x+b\quad\quad (2)\]
Then comparing equations \((1)\) and \((2)\), we get that \(m_1=-\dfrac{a}{b}\). And from the above theory of slope we know that the product of slopes of perpendicular lines is \(-1\).
\[\implies m_1 · m_2=-1\]
\[\begin{align} \implies m_2&=-\dfrac{1}{m_1}=\\&=-\dfrac{1}{-\frac{a}{b}}=\\&=\dfrac{b}{a}\\\\ \therefore m_2&=\dfrac{b}{a} \end{align}\]
Hence, from the given equation of line \(ax+by+c=0\), we can calculate the slopes of the perpendicular lines using the formula \(m_1=-\dfrac{a}{b}\), \(m_2=\dfrac{b}{a}\).
Suppose a line \(5x+3y+7=0\) is given. Find the slope for the line perpendicular to the given line.
Solution:
It is given that \(5x+3y+7=0\). Now comparing it with the general equation of line \(ax+by+c=0\), we get \(a=5\), \(b=3\), \(c=7\).
Now we use the above formula to calculate the slope.
\[\begin{align}\implies m_1&=-\dfrac{a}{b}=\\\\&=-\dfrac{5}{3}\end{align}\]
Now using the above-mentioned formula in the explanation, the slope of the perpendicular line is,
\[\begin{align}\implies m_2&=-\dfrac{b}{a}=\\\\&=-\dfrac{3}{5}\end{align}\]
Hence, the slope for the line perpendicular to \(5x+3y+7=0\) is \(m_2=\dfrac{3}{5}\).
Perpendicular line equation
The perpendicular line equation can be derived from the equation of a line that is written in the form \(y=mx+b\). We studied, that the slopes of perpendicular lines are the negative reciprocal of each other. So, when writing equations of perpendicular lines, we need to ensure that the slopes of each line when multiplied together get \(-1\).
If we want to find an equation for a line perpendicular to another line, we must take the negative reciprocal of that line’s slope. This value will be your value for \(m\) in the equation. The y-intercept can be anything, as a line can have infinitely many perpendicular lines that intersect with it. So, unless the question states otherwise, you can use any value for \(b\).
Find the equation of a line passing through the point \((0,2)\) such that it is perpendicular to the line \(y=2x-1\).
Solution:
First, we find the slope for the perpendicular line. Here, the equation for one line is given \(y=2x-1\). Comparing it with the general equation of line \(y=mx+b\), we get \(m_1=2\).
Now we take the negative reciprocal of the above slope to find the slope for the other line.
\[\implies m_2=-\dfrac{1}{m_1}\]
\[\implies m_2=-\dfrac{1}{2}\]
Now it is mentioned in the question that the other line passes through the point \((0,2)\). So the y-intercept for this line will be,
\[y=mx+b\]
\[\begin{align} &\implies y=\left(-\dfrac{1}{2}\right)x+b\\&\implies 2y=-x+2b\\&\implies 2y+x=2b\\&\implies 2(2)+0=2b\quad \quad\quad \text{substitute point }(0,2)\\&\implies 4=2b\\ &\therefore b=2 \end{align}\]
Now finally we substitute all the obtained values in the equation of the line.
\[y=mx+b\]
\[\therefore y=-\dfrac{1}{2}x+2\]
Graphically, we can show the obtained perpendicular lines as below.
Perpendicular lines graph, StudySmarter Originals
Perpendicular lines example
Let us take a look at some examples of perpendicular lines.
Check if the given lines are perpendicular or not.
Line 1: \(4x-y-5=0\), Line 2: \(x+4y+1=0\).
Solution:
To check if the given lines are perpendicular, we will see if the product of the slopes is \(-1\) or not. So comparing the given equations of line \(4x-y-5=0\), \(x+4y+1=0\) with the general form \(ax+by+c=0\).
\[\implies a_1=4,\quad b_1=-1,\quad c_1=-5;\quad a_2=1,\quad b_2=4,\quad c_2=1\]
Now we use the formula to calculate the slope for perpendicular lines. Therefore, for the line 1, we get
\[\implies m_1=-\dfrac{a_1}{b_1}=-\dfrac{4}{(-1)}=\dfrac{4}{1}=4\]
And for the line 2, the slope is
\[\implies m_2=-\dfrac{a_2}{b_2}=-\dfrac{1}{4}\]
Here \(m_1=4\), \(m_2=-\dfrac{1}{4}\) are negative reciprocal of each other. So, the product of both of them is
\[m_1 ·m_2=4\times \left(-\dfrac{1}{4}\right)=-1\]
Hence, both the given lines are perpendicular to each other.
Find the equation of the line if it passes through the point \((0,1)\) and is perpendicular to another line \(x+y=6\).
Solution:
Here, the equation for the first line is given as \(x+y=6\). And the second line passes through the point \((0,1)\). Now we simplify the given equation of line such that it looks similar to the form \(y=mx+b\).
\[\implies x+y=6\]
\[\begin{align} \implies y&=6-x\\ &=-x+6\\&=(-1)x+6\\\therefore \,y&=-1x+6 \end{align}\]
So, comparing this obtained equation with the general form of the line from above, we get \(m_1=-1\), \(b_1=6\) for the first line. Now, to find the slope of the second line, we know that it is a negative reciprocal of the slope of the first line.
\[\begin{align}\implies m_2&=-\dfrac{1}{m_1}\\&=-\dfrac{1}{(-1)}\\ \therefore m_2&=1\end{align}\]
And as the second line passes through the point \((0,1)\), the y-intercept is,
\[y=m_2 x+b_2\]
\[\begin{align}\implies y&=(1)x+b_2\\ \implies y&=x+b_2\\ \implies 1&=0+b_2\quad \quad\quad \text{substitute point (0,1)}\\ \therefore b_2&=1\end{align}\]
So putting all the obtained values in the general form of line, we get,
\[\begin{align}y&=m_2x+b_2\\&=1x+1\\&=x-1\end{align}\]
The equation of the line which is perpendicular to \(x+y=6\) and passing through \((0,1)\) is \(y=x+1\).
Perpendicular Lines - Key takeaways
- Two distinct straight lines which intersect at \(90º\) are called perpendicular lines.
- The slope of the perpendicular lines are negative reciprocal of each other.
- The slopes of the perpendicular lines using the formula \(m_1=-\dfrac{a}{b}\), \(m_2=\dfrac{b}{a}\).
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