DISTANCE FROM A POINT TO THE GIVEN LINE

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Consider a point P(x1, y1) and a line ax + by + c = 0 in the line co – ordinate axes. clip_image004

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The coordinates of points A and B are \left( { - \frac{c}{a},0} \right) and\left( {0, - \frac{c}{b}} \right) .

Area of \Delta PAB = \left| {\frac{1}{2}\left[ {{x_1}\left( {0 + \frac{c}{b}} \right) - \frac{c}{a}\left( { - \frac{c}{b} - {y_1}} \right) + 0} \right]} \right|

  \Rightarrow \,\frac{1}{2} \times AB \times PM = \left| {\frac{1}{2}\left[ {\frac{{c{x_1}}}{b} + \frac{{{c^2}}}{{ab}} + \frac{{c{y_1}}}{a}} \right]} \right|

 \Rightarrow \,\sqrt {\frac{{{c^2}}}{{{a^2}}} + \frac{{{c^2}}}{{{b^2}}}} \, \times \,PM = \left| {\frac{{ac{x_1} + {c^2} + bc{y_1}}}{{ab}}} \right|

 \Rightarrow \,\frac{c}{{ab}}\sqrt {{a^2} + {b^2}}  \times \,PM = \left| {\frac{c}{{ab}}\left( {a{x_1} + b{y_1} + c} \right)} \right|

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

Example: Find the distance from the point (1, 2) to the line 2x + 3y = 5 .

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

 \Rightarrow \,\,PM = \frac{{\left| {1 \times 2 + 2 \times 3 - 5} \right|}}{{\sqrt {{2^2} + {3^2}} }}\,

 \Rightarrow \,\,PM\, = \,\frac{3}{{\sqrt {13} }}

clip_image024 The required distance = \frac{3}{{13}}\sqrt {13} units.

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