# 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.