# Distance of a Point from a Line

The distance $d$ of the point $A\left( {{x_1},{y_1}} \right)$ from the line $ax + by + c = 0$ is given by

Let $\alpha$ be the inclination of the line as show in the given diagram, then its slope is given as

Squaring both sides of the above value of slope, we get

Draw a perpendicular from point $A$ to X-axis intersecting the line at point $P$. Since the abscissa of $P$ is same as that of $A$, so coordinates of point $P$ are $\left( {{x_1},{y_2}} \right)$, ${y_2} \ne {y_1}$. In the triangle $ALP$, $m\angle MAP = \alpha$ and $m\angle AMP = {90^ \circ }$, so

Since the distance $d$ should be positive, so we must take the modulus of the right side of above equation, i.e.

Putting the value of $\left| {\cos \alpha } \right|$ from equation (i) in equation (ii), we get

Since the point $P\left( {{x_1},{y_2}} \right)$ lies on the line $ax + by + c = 0$, so

Putting this value in equation (iii), we get

Example: Find the distance of the point $\left( {2, - 3} \right)$ from the line $7x - 5y + 9 = 0$.
Since the given point is $\left( {2, - 3} \right)$, so ${x_1} = 2,\,\,{y_1} = - 3$
Since the given line is $7x - 5y + 9 = 0$, so $a = 7,\,\,b = - 5,\,\,c = 9$
The required distance is