# 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.

Yash

January 21@ 11:00 amthe point (4,0) lies below the line given by the coordinates (2,4) and (4,1). But this case is not satisfied by the theory given above.

Kartik

June 28@ 6:53 amThe point is below the line.

The eq of the line given by these two points will be : 3x-2y-10=0=f(x,y)

Now, f(4,0) = 3*4-2*0-10 = 2 > 0

But, b<0 ( b= -2)

Since both have opposite sign, the given point is below the line. The theory above is correct.