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I was given a test yesterday, a test which unfortunately I was unable to study to. In it was a question that was too hard that it became our homework for the whole week.

It says that on lines $4x+3y-6=0$ and $2x+3y+4=0$, we need to find 2 points on either of the two lines that is a distance of 2 units away from the other line.

I tried tackling the problem repeatedly but to no avail, although I did get their point of intersection which is $(34,-24)$. A little help please?

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Are you looking for two points, one on each line, that are $2$ units apart? Since you have the intersection point, $(34,-24)$, the easiest way is to find a point on one of the lines that is $2$ units from it. Is that allowed? On the second line (I chose that one because the $3-4-5$ triangle will keep the result rational) we can move $+4$ units in $x$ and $-3$ in $y$ and stay on the line. This is a distance of $5$. Since we want a distance of $2$, we multiply by $\frac 25$, getting a change in $x$ of $\frac 85$ and a change in $y$ of $\frac {-6}5$. The point is then $(35\frac 35,-25\frac 15)$

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That did it thanks. Never thought of using the triangle to find the right points on the line. –  Aldon Oct 11 '12 at 22:09
    
I don’t think that @Ross’s analysis of the problem is what was meant in its statement. It looks to me rather as if one is being asked to find a point $P$ on the first line that is $2$ units from the other line. This distance is measured along the perpendicular from $P$ to the line. –  Lubin Oct 11 '12 at 22:15
    
@Lubin: Américo Tavares took it the same way you did and posted a nice answer, but deleted it. I was not sure I had it right, either. –  Ross Millikan Oct 12 '12 at 0:06
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