# Find the point on the line 6x + 3y + 2 = 0 that is closest to the point (-2,-1).

## Question:

Find the point on the line {eq}6x + 3y + 2 = 0 {/eq} that is closest to the point {eq}(-2,-1) {/eq}.

## Distance Between a Point and a Line:

When we know the equation of a line {eq}L:\,\,ax + by + c = 0 {/eq} and a point {eq}P\left( {{x_0},{y_0}} \right) {/eq} outside the line we are able to apply the corresponding formula to derive the distance between the point {eq}P {/eq} and the line {eq}L {/eq}: {eq}d\left( {P,L} \right) = \frac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }} {/eq}.

## Answer and Explanation:

{eq}\eqalign{ & {\text{If we have that the equation of the line }}\,L{\text{ is }}\,ax + by + c = 0{\text{ and a point }}\,{P_0}\left( {{x_0},{y_0}} \right){\text{ }} \cr & {\text{outside the line}}{\text{, then}}{\text{, the shortest distance from the point }}\,{P_0}{\text{ to the line }}L{\text{ is given by: }} \cr & \,\,\,\,d\left( {P,L} \right) = \frac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }} \cr & {\text{Now}}{\text{, in this particular case we have the point }}P\left( {{x_0},{y_0}} \right) = \left( { - 2, - 1} \right){\text{ and the line}} \cr & 6x + 3y + 2 = 0.{\text{ So}}{\text{, by replacing }}a = 6,\,\,b = 3,\,\,c = 2{\text{:}} \cr & \,\,\,\,d\left( {P,L} \right) = \frac{{\left| {6 \cdot \left( { - 2} \right) + 3 \cdot \left( { - 1} \right) + 2} \right|}}{{\sqrt {{6^2} + {3^2}} }} \cr & \,\,\,\,d\left( {P,L} \right) = \frac{{\left| { - 12 - 3 + 2} \right|}}{{\sqrt {36 + 9} }} = \frac{{13}}{{\sqrt {45} }} = \boxed{1.9379} \cr & {\text{Therefore}}{\text{, the closest distance from the point }}\,{P_0}{\text{ to the line }}L{\text{ is }}\,\boxed{d\left( {P,L} \right) = 1.9379} \cr} {/eq}

#### Learn more about this topic:

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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