# Position of Point with Respect to a Line

The general equation or standard equation of a straight line is given by
$ax + by + c = 0\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$ Let $A\left( {{x_1},{y_1}} \right)$ be a point which does not lie on the line (i). Now draw a perpendicular line from $A$ to the X-axis through the point $P$ as shown in the given diagram. It is clear from the diagram that ${x_1}$ is the abscissa of the point $P$. If $y$ is the ordinate of $P$, then $\left( {{x_1},y} \right)$ are the coordinates of $P$. Since $P$ lies on the line (i), it must satisfy the equation of the line, i.e.
$\begin{gathered} a{x_1} + by + c = 0 \\ \Rightarrow by = – a{x_1} – c \\ \Rightarrow y = – \frac{{a{x_1} + c}}{b}\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered}$

Next we consider the difference ${y_1} – y$, i.e.
$\begin{gathered} {y_1} – y = {y_1} – \left( { – \frac{{a{x_1} + c}}{b}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {y_1} + \frac{{a{x_1} + c}}{b} = \frac{{b{y_1} + a{x_1} + c}}{b} \\ \Rightarrow {y_1} – y = \frac{{a{x_1} + b{y_1} + c}}{b}\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right) \\ \end{gathered}$

(a) If the point $A$ is above the line, then ${y_1} – y > 0$. From equation (iii), we note that ${y_1} – y > 0$ only if $\frac{{a{x_1} + b{y_1} + c}}{b} > 0$ . But $\frac{{a{x_1} + b{y_1} + c}}{b} > 0$ if $a{x_1} + b{y_1} + c > 0$ and $b > 0$ or $\frac{{a{x_1} + b{y_1} + c}}{b} > 0$ if $a{x_1} + b{y_1} + c < 0$ and $b < 0$.

We conclude that the point $A\left( {{x_1},{y_1}} \right)$ is above the line $ax + by + c = 0$ if
(i) $a{x_1} + b{y_1} + c > 0$ and $b > 0$
(ii) $a{x_1} + b{y_1} + c < 0$ and $b < 0$

(b) If the point $A$ is below the line, then ${y_1} – y < 0$. From equation (iii), we note that ${y_1} – y < 0$ only if $\frac{{a{x_1} + b{y_1} + c}}{b} < 0$ . But $\frac{{a{x_1} + b{y_1} + c}}{b} < 0$ if $a{x_1} + b{y_1} + c < 0$ and $b > 0$ or $\frac{{a{x_1} + b{y_1} + c}}{b} < 0$ if $a{x_1} + b{y_1} + c > 0$ and $b < 0$.

We conclude that the point $A\left( {{x_1},{y_1}} \right)$ is below the line $ax + by + c = 0$ if
(i) $a{x_1} + b{y_1} + c < 0$ and $b > 0$
(ii) $a{x_1} + b{y_1} + c > 0$ and $b < 0$

NOTE: The point $A\left( {{x_1},{y_1}} \right)$ will be on the line $ax + by + c = 0$ if $a{x_1} + b{y_1} + c = 0$.

Example: Determine whether the point $\left( {1,3} \right)$ lies below or above the line $3x – 2y + 1 = 0$.
Comparing the given line $3x – 2y + 1 = 0$ with the general equation of line $ax + by + c = 0$, we have $a = 3$, $b = – 2 < 0$ and $c = 1$.

Since $\left( {1,3} \right)$ is the given point, then ${x_1} = 1,\,{y_1} = 3$.

Now:
$a{x_1} + b{y_1} + c = 3\left( 1 \right) + \left( { – 2} \right)\left( 3 \right) + 1 = – 2 < 0$

Since $a{x_1} + b{y_1} + c < 0$ and $b < 0$, the given point lies above the line.