Question 359558
1.{{{y=-2x+1}}}
The closest distance is formed by the line perpendicular to {{{y=-2x+1}}} that goes through the point (-4,-1).
To find the perpendicular line, use the slope since perpendicular lines have slopes that are negative reciprocals.
{{{m[1]*m[2]=-1}}}
{{{-2*m[2]=-1}}}
{{{m[2]=1/2}}}
Now use the point slope form of a line, {{{y-y[p]=m(x-x[p])}}}
{{{y-(-1)=(1/2)(x-(-4))}}}
{{{y+1=(1/2)(x+4)}}}
{{{2y+2=x+4}}}
2.{{{-x+2y=2}}}
Now that you have both lines, find the intersection of the two lines. 
The distance you're after is the distance from point (-4,-1) to the intersection point.
Substitute eq. 1 into eq. 2 and solve for {{{x}}}.
{{{-x+2(-2x+1)=2}}}
{{{-x-4x+2=2}}}
{{{-5x=0}}}
{{{x=0}}}
Then from eq. 1,
{{{y=-2(0)+1}}}
{{{y=1}}}
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{{{drawing(300,300,-5,5,-5,5,grid(1),circle(-4,-1,0.2),circle(0,1,0.2),graph(300,300,-5,5,-5,5,-2x+1,(2+x)/2))}}}
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Now use the distance formula to find the distance from (-4,-1) to (0,1).
{{{D^2=(x[2]-x[1])^2+(y[2]-y[1])^2}}}
{{{D^2=(0-(-4))^2+(1-(-1))^2}}}
{{{D^2=(4)^2+(2)^2}}}
{{{D^2=20}}}
{{{highlight(D=2*sqrt(5))}}} or approximately,
{{{D=4.47}}}