Question 849870
Do you mean the minimum distance to the line? 
If so, the minimum distance is a perpendicular line going through the point.
A perpendicular line would have the form,
{{{y=-(1/2)x+b}}}
Use the point to solve for {{{b}}}.
{{{-3=-(1/2)(6)+b}}}
{{{-3=-3+b}}}
{{{b=0}}}
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Find the intersection of the two lines.
{{{y=-(1/2)x}}}
{{{y=2x-5}}}
Set them equal to each other.
{{{-(1/2)x=2x-5}}}
{{{-x=4x-10}}}
{{{-5x=-10}}}
{{{x=2}}}
Then from either equation, solve for {{{y}}}.
{{{y=-(1/2)(2)=-1}}}
Now find the distance from (6,-3) to (2,-1) using the distance formula,
{{{D^2=(x[2]-x[1])^2+(y[2]-y[1])^2}}}
{{{D^2=(2-6)^2+(-3-(-1))^2}}}
{{{D^2=(-4)^2+(-2)^2}}}
{{{D^2=16+4}}}
{{{D^2=20}}}
{{{highlight(D=2sqrt(5))}}}
{{{drawing(300,300,-2,8,-8,2,grid(1),
circle(6,-3,0.2),
circle(2,-1,0.2),
line(6,-3,2,-1),
graph(300,300,-2,8,-8,2,2x-5,-x/2))}}}