Perpendicular Lines

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 \(m_2=\dfrac{3}{5}\).

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

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 \(y=x+1\).