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Use a scalar projection to show that the distance from a point P(x0,y0) to the line L : ax + by +...

Question:

Use a scalar projection to show that the distance from a point {eq}P(x0,y0){/eq} to the line L : {eq}ax + by + c = 0{/eq} is,

{eq}d(P,L) = \frac{|ax0 + by0 + c| }{ \sqrt a2 + b2}{/eq} . Use this formula to find the distance from point {eq}(-2,3){/eq} to the line {eq}6x-8y-1=0{/eq}

Distance Between a Point and a Line and Scalar Projection:

When we know the direction vector of a line and a point {eq}{P_0}\left( {{x_0},{y_0}} \right) {/eq} outside the line we can set a normal vector {eq}{\vec n} {/eq} to the line that goes from point {eq}{P_0} {/eq} outside the line. Then we can calculate the scalar projection of any vector that goes from point {eq}{P_0} {/eq} to any point {eq}{P_1}\left( {x,y} \right) {/eq} on the line onto the normal vector {eq}{\vec n} {/eq}, so we get the distance between {eq}{P_0} {/eq} and the line: {eq}\,d = \,com{p_{\vec n}}\overrightarrow {{P_0}{P_1}} = \frac{{\vec n \cdot \overrightarrow {{P_0}{P_1}} }}{{\left| {\vec n} \right|}} {/eq}.

Answer and Explanation:

{eq}\eqalign{ & {\text{We have that the equation of the line is }}\,ax + by + c = 0.{\text{ Then its direction vector }} \cr & {\text{(vector parallel to the line) is }}\vec v = \left\langle {b, - a} \right\rangle \cr & {\text{So}}{\text{, two vector }}{\mkern 1mu} \vec v = \left\langle {{v_x},{v_y}} \right\rangle {\mkern 1mu} {\text{ and }}{\mkern 1mu} \vec n{\text{ are perpendicular if }}{\mkern 1mu} \vec n = \left\langle {{v_y}, - {v_x}} \right\rangle {\text{, thus the normal}} \cr & {\text{vector to the line is: }}\,\vec n = \left\langle {a,b} \right\rangle {\text{.}} \cr & {\text{Now we establish a vector between the point }}\,{P_0}\left( {{x_0},{y_0}} \right){\text{ outside the line and a }} \cr & {\text{point }}\,{P_1}\left( {x,y} \right){\text{ contained in the line:}} \cr & \,\,\,\overrightarrow {{P_0}{P_1}} = \left\langle {x - {x_0},y - {y_0}} \right\rangle \cr & {\text{Let's find the scalar projection of }}\,\overrightarrow {{P_0}{P_1}} = \left\langle {x - {x_0},y - {y_0}} \right\rangle \,{\text{ onto }}\,\vec n = \left\langle {a,b} \right\rangle {\text{, so}}{\text{, the }} \cr & {\text{scalar projection is given by:}} \cr & \,\,d = \,com{p_{\vec n}}\overrightarrow {{P_0}{P_1}} = \frac{{\vec n \cdot \overrightarrow {{P_0}{P_1}} }}{{\left| {\vec n} \right|}} = \frac{{\left\langle {a,b} \right\rangle \cdot \left\langle {x - {x_0},y - {y_0}} \right\rangle }}{{\left| {\left\langle {a,b} \right\rangle } \right|}} \cr & \,\,d = \frac{{a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right)}}{{\sqrt {{a^2} + {b^2}} }} \cr & \,\,d = \frac{{ax - a{x_0} + by - b{y_0}}}{{\sqrt {{a^2} + {b^2}} }} \cr & \,\,d = \frac{{ - a{x_0} - b{y_0} + ax + by}}{{\sqrt {{a^2} + {b^2}} }} \cr & {\text{Simplifying using }}\,ax + by + c = 0\,\,\, \Rightarrow - c = ax + by{\text{:}} \cr & \,\,d = \frac{{ - a{x_0} - b{y_0} - c}}{{\sqrt {{a^2} + {b^2}} }} \cr & {\text{By definition the distance can only be positive:}} \cr & \,\,d = \frac{{\left| { - \left( {a{x_0} + b{y_0} + c} \right)} \right|}}{{\sqrt {{a^2} + {b^2}} }} \cr & {\text{Therefore:}} \cr & \,\,\boxed{d\left( {P,L} \right) = \frac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}} \cr & \cr & {\text{Now}}{\text{, in this particular case we have the point }}P\left( {{x_0},{y_0}} \right) = \left( { - 2,3} \right){\text{ and the line}} \cr & 6x - 8y - 1 = 0.{\text{ So}}{\text{, by replacing:}} \cr & \,\,\,\,d\left( {P,L} \right) = \frac{{\left| {6 \cdot \left( { - 2} \right) - 8 \cdot 3 - 1} \right|}}{{\sqrt {{6^2} + {8^2}} }} \cr & \,\,\,\,d\left( {P,L} \right) = \frac{{\left| { - 12 - 24 - 1} \right|}}{{\sqrt {36 + 64} }} = \frac{{37}}{{\sqrt {100} }} = \boxed{3.7} \cr} {/eq}


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How to Find the Distance Between a Point & a Line

from FTCE Mathematics 6-12 (026): Practice & Study Guide

Chapter 30 / Lesson 5
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