# Find the point on the line -2x+4y-5=0 which is closest to the point (0,-2).

## Question:

Find the point on the line {eq}-2x+4y-5=0 {/eq} which is closest to the point {eq}(0,-2) {/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}.

{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( {0, - 2} \right){\text{ and the line}} \cr & - 2x + 4y - 5 = 0.{\text{ So}}{\text{, by replacing:}} \cr & \,\,\,\,d\left( {P,L} \right) = \frac{{\left| { - 2 \cdot 0 + 4 \cdot \left( { - 2} \right) - 5} \right|}}{{\sqrt {{{\left( { - 2} \right)}^2} + {4^2}} }} \cr & \,\,\,\,d\left( {P,L} \right) = \frac{{\left| {0 - 8 - 5} \right|}}{{\sqrt {4 + 16} }} = \frac{{13}}{{\sqrt {20} }} = \boxed{2.9069} \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) = 2.9069} \cr} {/eq} 