# Equation of a Line Perpendicular to Another Line

Consider that we have the given equation of a line, and this given line is perpendicular to another line which passes through the given point, and we must find the required equation of the line with the help of the given line.

Now we follow the following procedure with the help of an example.

__Example__**:** Find the equation of a straight line passing through the point $$\left( {1,2} \right)$$ and is perpendicular to another given line whose equation is $$2x – 3y + 5 = 0$$.

First find the slope of the given line by comparing the slope intercept form of a line as follows:

\[\begin{gathered} 2x – 3y + 5 = 0 \\ \Rightarrow 3y = 2x + 5 \\ \Rightarrow y = \frac{2}{3}x + \frac{5}{3} \\ \end{gathered} \]

Compare this with the slope intercept form to find the slope of the given line $$y = mx + c$$.

Now the slope of the given line is $$m = \frac{2}{3}$$, since the given line is perpendicular to the required line under the condition of perpendicular lines $${m_1} \times {m_2} = – 1$$.

Now the slope of the required line is $$m = – \frac{3}{2}$$, since the required line passes through the given point $$\left( {1,2} \right)$$. To find the equation of the required line use the slope point form, which is given by:

\[\begin{gathered} y – {y_1} = m\left( {x – {x_1}} \right) \\ \Rightarrow y – 2 = – \frac{3}{2}\left( {x – 1} \right) \\ \Rightarrow 2y – 4 = – 3x + 3 \\ \Rightarrow 3x + 2y – 7 = 0 \\ \end{gathered} \]

This is the equation of a straight line perpendicular to the line $$2x – 3y + 5 = 0$$. In this calculated equation we observe that the coefficients of $$x$$ and $$y$$ cross each other with a negative sign between them.