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

d =  \frac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}



distance-point-from-line

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

\tan  \alpha  =   - \frac{a}{b}


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

\begin{gathered} {\tan ^2}\alpha  = \frac{{{a^2}}}{{{b^2}}} \\ \Rightarrow {\sec ^2}\alpha  - 1 =  \frac{{{a^2}}}{{{b^2}}}\,\, \Rightarrow {\sec ^2}\alpha  = \frac{{{a^2}}}{{{b^2}}} + 1 \\ \Rightarrow {\sec ^2}\alpha  = \frac{{{a^2} + {b^2}}}{{{b^2}}}\,\,  \Rightarrow {\cos ^2}\alpha  =  \frac{{{b^2}}}{{{a^2} + {b^2}}} \\ \Rightarrow \cos \alpha  =  \pm  \frac{b}{{\sqrt {{a^2} + {b^2}} }} \\ \Rightarrow \left| {\cos \alpha } \right| =  \frac{{\left| b \right|}}{{\sqrt {{a^2} + {b^2}} }}\,\,\,\,{\text{ -  -  -  }}\left( {\text{i}} \right) \\ \end{gathered}


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

\frac{{AM}}{{AP}}  = \cos \alpha


\begin{gathered} \Rightarrow AM = AP\cos \alpha \\ \Rightarrow d = \left( {AL - PL} \right)\cos  \alpha \,\,\,\,\,\,\,\,\,\,\because AM = d,\,\,AM = AL - PL \\ \Rightarrow d = \left( {{y_1} - {y_2}}  \right)\cos \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\because AL = {y_1},\,PL = {y_2} \\ \end{gathered}


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

d =  \left| {{y_1} - {y_2}} \right|\left| {\cos \alpha } \right|\,\,\,\,{\text{  -  -   - }}\left( {{\text{ii}}} \right)


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

d =  \left| {{y_1} - {y_2}} \right|\frac{{\left| b \right|}}{{\sqrt {{a^2} + {b^2}}  }}\,\,\,\,{\text{ -  -  - }}\left( {{\text{iii}}} \right)


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

\begin{gathered} a{x_1} + b{y_2} + c = 0 \\ \Rightarrow b{y_2} =  - a{x_1} - c \\ \Rightarrow {y_2} = \frac{{ - a{x_1} -  c}}{b} \\ \end{gathered}


Putting this value in equation (iii), we get

d =  \left| {{y_1} - \frac{{ - a{x_1} - c}}{b}} \right|\frac{{\left| b \right|}}{{\sqrt  {{a^2} + {b^2}} }}\,\,


\begin{gathered} d = \left| {\frac{{b{y_1} + a{x_1} + c}}{b}}  \right|\frac{{\left| b \right|}}{{\sqrt {{a^2} + {b^2}} }} = \frac{{\left|  {a{x_1} + b{y_1} + c} \right|}}{{\left| b \right|}}\frac{{\left| b  \right|}}{{\sqrt {{a^2} + {b^2}} }} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \Rightarrow \boxed{d = \frac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt  {{a^2} + {b^2}} }}} \\ \end{gathered}


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

d = \frac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt  {{a^2} + {b^2}} }} = \frac{{\left| {7\left( 2 \right) - 5\left( { - 3} \right)  + 9} \right|}}{{\sqrt {{7^2} + {{\left( { - 3} \right)}^2}} }} =  \frac{{38}}{{\sqrt {74} }}

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