Question 1043689
Find the slope of each line.
{{{2x-y+3=0}}}
{{{y[1]=2x+3}}}
{{{m[1]=2}}}
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{{{x-4y-7=0}}}
{{{4y=x-7}}}
{{{y[2]=x/4-7/4}}}
{{{m[2]=1/4}}}
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The perpendicular line from the origin to each line has a slope that is the negative reciprocal of the given slope.
{{{m[1p]=-1/2}}}
{{{m[2p]=-4}}}
So then each line goes through the origin so the equation of each line is,
{{{y-0=-1/2(x-0)}}}
{{{y[1p]=-(1/2)x}}}

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{{{y-0=-4(x-0)}}}
{{{y[2p]=-4x}}}
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Find the intersection point of each line,
{{{2x+3=-(1/2)x}}}
{{{(5/2)x=-3}}}
{{{x=-6/5}}}
So then,
{{{y=-(1/2)(6/5)}}}
{{{y=3/5}}}
({{{-6/5}}},{{{3/5}}})
and
{{{x/4-7/4=-4x}}}
{{{(17/4)x=7/4}}}
{{{x=7/17}}}
Then,
{{{y=-4(7/17)}}}
{{{y=-28/17}}}
({{{7/17}}},{{{-28/17}}})
So then calculate the distance from the origin to the respective intersection points,
{{{D[1]^2=(-6/5-0)^2+(3/5-0)^2}}}
{{{D[1]^2=(36/25)+(9/25)}}}
{{{D[1]^2=45/25}}}
{{{D[1]=sqrt(45)/5}}}
{{{D[1]=(3/5)sqrt(5)}}}
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{{{D[2]^2=(7/17-0)^2+(-28/17-0)^2}}}
{{{D[2]^2=(49+784)/17^2}}}
{{{D[2]=sqrt(833)/17}}}
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Here's a graphical representation of the problem.
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*[illustration gr2.JPG].