$ax+by+c = 0 \,\,\,$ (or) $\,\,\, ax+by = c$

Geometrically, the straight lines are possibly appeared in five different cases in two dimensional space and they are expressed in algebraic equation form mathematically. The five types of equations of straight lines can be represented generally by a standard form. It is called as general form or standard form of a straight line.

It is usually written as $ax+by+c = 0$ or $ax+by = c$, where $a$, $b$ and $c$ are constants and $x$ and $y$ are variables.

Now, learn the way of transforming equations of straight lines in general form.

$y \,=\, mx+c$ is a slope and y-intercept form a straight line. It can be written in standard form as $mx-y+c = 0$. It is in the form of $ax+by+c = 0$. Therefore, $a = m$, $b = -1$ and $c = c$ by the comparing them.

Similarly, it can also be written as $mx-y = -c$ and it is in the form of $ax+by = c$. In this case, it is known that $a = m$, $b = -1$ and $c = -c$.

$x \,=\, m’y+c$ is a slope and x-intercept form a straight line. It can be written in general form as $x-m’y-c = 0$ and it is in the form of $ax+by+c = 0$. By comparing them, It is known that $a = 1$, $b = -m’$ and $c = -c$

If, it is written as $x-m’y = c$, then it is in the form of $ax+by = c$. So, it is clear that $a = 1$, $b = -m’$ and $c = c$.

$y \,=\, mx$

$\dfrac{x}{a}+\dfrac{y}{b} \,=\, 1$

$y-y_{1} \,=\, m(x-x_{1})$

$y-y_{1} \,=\, \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$

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