Question 374990
The shortest distance between a line and a point is a line perpendicular to the first line going through the point.
Perpendicular lines have slopes that are negative reciprocals.
{{{m[1]*m[2]=-1}}}
{{{(1/2)*m[2]=-1}}}
{{{m[2]=-2}}}
Then use the slope-intercept form of a line,
{{{y=mx+b}}}
{{{y=-2x+b}}}
Use the point to solve for {{{b}}}.
{{{-1=-2(2)+b}}}
{{{b=-1+4}}}
{{{b=3}}}
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{{{y=-2x+3}}}
Find the intersection point between the two lines.
{{{-2x+3=(1/2)x+4}}}
{{{-(5/2)x=1}}}
{{{x=-2/5}}}
Then,
{{{y=-2(-2/5)+3}}}
{{{y=4/5+3}}}
{{{y=19/5}}}
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Use the distance formula to find the distance from ({{{2}}},{{{-1}}}) to ({{{-2/5}}},{{{19/5}}}).
{{{D^2=(x[2]-x[1])^2+(y[2]-y[1])^2}}}
{{{D^2=(-2/5-2)^2+(19/5-(-1))^2}}}
{{{D^2=(-2/5-10/5)^2+(19/5+5/5)^2}}}
{{{D^2=(-12/5)^2+(24/5)^2}}}
{{{D^2=(144/25)+(576/5)}}}
{{{D^2=720/25}}}
{{{highlight(D=(12/5)*sqrt(5))}}}

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{{{drawing(300,300,-5,5,-5,5,grid(1),circle(-2/5,19/5,0.2),circle(2,-1,0.2),graph(300,300,-5,5,-5,5,0,(1/2)x+4,-2x+3))}}}