Distance Between a Line and a Point Using VectorsDate: 09/25/1999 at 17:08:46 From: Ian Bui Subject: Distance between line and point using vectors How do you use scalar vector projections to prove that the distance between the line ax+by+c = 0 and point (x1,y1) is |ax1+by1+c|/sqrt(a^2+b^2) I tried choosing an arbitrary point on the line, but with each step, things got more complex. Date: 09/25/1999 at 19:46:50 From: Doctor Anthony Subject: Re: Distance between line and point using vectors The equation ax + by + c = 0 can be written: y = -(a/b)x - (c/a) So the gradient of line is -(a/b). This means that the cosine of the angle between the line and the x axis is -b/sqrt(a^2+b^2) and the cosine of the angle between the line and the y axis is a/sqrt(a^2+b^2). So the direction ratios of the line are (-b,a). If we have a line perpendicular to this, its direction ratios will be (a,b) since (-ab) + (ab) = 0. So the perpendicular line through (x1,y1) is: x-x1 y-y1 ---- = ---- = t/sqrt(a^2+b^2) where t is a parameter. a b t will be the distance from the point (x1,y1) along the perpendicular. So x = x1 + a.t/sqrt(a^2+b^2) and y = y1 + b.t/sqrt(a^2+b^2) This meets the given line ax+by+c = 0 where: a(x1+a.t/sqrt(a^2+b^2)) + b(y1+b.t/sqrt(a^2+b^2)) + c = 0 ax1 + by1 + c + t(a^2+b^2)/sqrt(a^2+b^2) + c = 0 ax1 + by1 + c + t.sqrt(a^2+b^2) = 0 so t.sqrt(a^2+b^2) = -(ax1+by1+c) t = -(ax1+by1+c)/sqrt(a^2+b^2) Therefore the perpendicular distance from (x1,y1) to the line ax+by+c = 0 is: ax1 + by1 + c |t| = ------------- sqrt(a^2+b^2) - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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