Perpendicular DistanceDate: 04/10/98 at 13:12:25 From: Andromeda Albester Subject: Geometry Coordinates Dear Dr. Maths: Could you please prove the following formula: The distance (d) from a 90 degree line dot (h,k) to ax + by + c = 0 is given by: ah+bk+c --------------- (a^2+b^2)^(1/2) I can't prove it at all. When asked my teachers, they just said it's beyond my level. Please help. Thanks. Date: 04/10/98 at 16:40:27 From: Doctor Rob Subject: Re: Geometry Coordinates I think you may be able to understand this proof. The line through (h,k), which is perpendicular to a*x + b*y + c = 0, is given by b*(x-h) - a*(y-k) = 0, I claim. You can easily check that it passes through (h,k), and its slope is b/a, and the slope of the original line is -a/b. Now oblique lines are perpendicular if and only if their slopes are negative reciprocals of each other, so these lines are perpendicular. See this by drawing the coordinate axes, and the line y = (b/a)*x, passing through the origin O and A(a,b). Then also draw the line y = -(a/b)*x, passing through the origin and B(b,-a). Let C = (a,0), D = (0,-a), and draw the triangles OAC and OBD. Show that they are congruent right triangles. Then angle AOC equals angle BOD. Last, show that the angle AOB equals angle COD, which is a right angle. I have not dealt with the special case where one line is horizontal, the other vertical, that is, one of a or b is zero. In that case, one line has slope 0, and the other line has no slope, but my claim, that the equation of the perpendicular line is as stated, still holds. (Check it out!) Now find the intersection of these two lines. When I did this, I found that the intersection was the point: (x0, y0) = ([b^2*h-a*b*k-a*c]/[a^2+b^2],[-a*b*h+a^2*k-b*c]/[a^2+b^2]) Then find the distance from that intersection to (h,k) using the distance formula. The distance is: d = sqrt[(h-x0)^2+(k-y0)^2] = sqrt[(h-[b^2*h-a*b*k-a*c]/[a^2+b^2])^2+ (k-[-a*b*h+a^2*k-b*c]/[a^2+b^2])^2] = sqrt[([a^2*h+a*b*k+a*c])^2+([a*b*h+b^2*k+b*c])^2] ------------------------------------------------- [a^2+b^2] = sqrt[a^2*(a*h+b*k+c)^2+b^2*(a*h+b*k+c)^2] ----------------------------------------- [a^2+b^2] = sqrt[(a*h+b*k+c)^2]*sqrt[a^2+b^2] --------------------------------- [a^2+b^2] = |a*h+b*k+c| ------------- sqrt(a^2+b^2) Can you follow this? -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/