From: Charlie Dickman <charlied@...>

Date: Thu, 8 Oct 1998 20:39:25 -0400

Date: Thu, 8 Oct 1998 20:39:25 -0400

>I wonder if anyone anyone can help me with this little dilemma that I must >solve to get an image processing program to work. > >My program finds 3 points in a picture which I have the x and y coordinates >of: Ax,Ay, Bx,By and Cx,Cy. I want to figure out where a fourth point would >be. I want to find the point on a line that passes through A and B that is >closest to C. If anyone knows the method to do this the help would be greatly >appreciated. > >I knew I should have paid more attention in algebra class in high school. > >Thanks for any help, >-Joe Lertola OK, here goes... You want the coordinates of D (Dx, Dy) which is closest to the line formed by A and B (let's call it AB). This point will be on AB and on the line perpendicular to AB through C (the perpendicular distance is always the shortest). Let M = (By-Ay)/(Bx-Ax). Then M is the slope of AB and -1/M is the slope of CD. Finding the coordinates of D goes like this... Dx = ((Cx+((M^2)*Ax)-(M*Ay)+(M*Cy))/((M^2)+1) and Dy = (M*(Dx-Ax))+Ay While there's a lot of algebra involved, the subject you missed is analytic geometry. Charlie Dickman charlied@...