Converting Linear Equations in Standard Form to Slope Point Form

The general equation or standard equation of a straight line is:
\[ax + by + c = 0\]

Where $$a$$ and $$b$$ are constants and either $$a \ne 0$$ or $$b \ne 0$$.

Convert the standard equation of line $$ax + by + c = 0$$ into the slope point form \[y – {y_1} = m\left( {x – {x_1}} \right)\]

If the standard form of the line passes through the point $$\left( {{x_1},{y_1}} \right)$$, then this point must satisfy the standard equation, i.e.
\[\begin{gathered} a{x_1} + b{y_1} + c = 0 \\ \Rightarrow b{y_1} = – a{x_1} – c \\ \Rightarrow {y_1} = – \frac{{\left( {a{x_1} + c} \right)}}{b} \\ \end{gathered} \]

Comparing the equation with the slope intercept form, the slope is $$m = – \frac{a}{b}$$. The slope point form of the line is $$y – {y_1} = m\left( {x – {x_1}} \right)$$.

Now we put the values of $$m$$ and $${y_1}$$ in the slope point form $$y – {y_1} = m\left( {x – {x_1}} \right)$$.
\[\begin{gathered} \Rightarrow y – \left( { – \frac{{a{x_1} + c}}{b}} \right) = – \frac{a}{b}\left( {x – {x_1}} \right) \\ \Rightarrow y + \frac{{a{x_1} + c}}{b} = – \frac{a}{b}x + \frac{a}{b}{x_1} \\ \Rightarrow y + \frac{a}{b}{x_1} + \frac{c}{b} = – \frac{a}{b}x + \frac{a}{b}{x_1} \\ \Rightarrow y = – \frac{a}{b}x + \frac{a}{b}{x_1} – \frac{a}{b}{x_1} – \frac{c}{b} \\ \Rightarrow y = – \frac{a}{b}x – \frac{c}{b} \\ \Rightarrow y = – \frac{a}{b}x – \frac{{ac}}{{ab}} \\ \Rightarrow y = – \frac{a}{b}\left( {x + \frac{c}{a}} \right) \\ \end{gathered} \]

This is the equation of a line in point-slope form transferred from its general form.