Coordinate GeometryDate: 09/23/98 at 16:32:56 From: Peter Subject: Coordinate geometry Given two known coordinate points in cartesian plane pt1, given by (x1, y1), and pt2, given by (x2, y2), how can I locate a third point pt3, given by (x3, y3), which is perpendicular to the line joining pt1 and pt2 and is of certain distance d from either pt1 or pt2? I've established two equations with two unknowns as following: m = (y2-y1)/(x2-x1), which implies (y3-y1)/(x3-x1) = -1/m (eqn. 1) (y3-y1)^2 + (x3-x1)^2 = d^2 (eqn. 2) Since x1, y1, x2, y2, and d are known, I can solve for x3 and y3 variables using either the substitution or the elimination method. There will also be two values for both x3 and y3, since the equation becomes a quadratic equation. To simplify the whole thing, is there a standard equation I can use to solve x3 and y3? Your response will be greatly appreciated. Date: 09/24/98 at 12:04:16 From: Doctor Peterson Subject: Re: Coordinate geometry Hi, Peter. Your first statement of the problem sounded as if you wanted a point on the perpendicular bisector of P1P2 at distance d from each of P1 and P2. But your equations tell me you want it a distance d from P1 along the perpendicular at P1, which is much easier. You get an easy quadratic with no linear term. It can also be solved geometrically: A +----+P2 | / B| / | /L |/ C +---------------+ P1 \ | \ |D d \ | +P3 The two triangles are similar, and: A = x2-x1 B = y2-y1 L = sqrt(A^2+B^2) C = x3-x1 D = y3-y1 so we get: C B D A --- = --- --- = --- d L d L from which we get: d(y2-y1) x3 = x1 + C = x1 + Bd/L = x1 +/- --------------------------- sqrt((x2-x1)^2 + (y2-y1)^2) d(x2-x1) y3 = y1 + D = y1 + Ad/L = y1 +/- --------------------------- sqrt((x2-x1)^2 + (y2-y1)^2) (The +or- allows the perpendicular to extend in either direction. Both are + for one point, - for the other.) If what you really wanted was a point on the perpendicular bisector (say, the center of a circle of radius d containing P1 and P2), you can apply this formula to the midpoint P4 of P1P2 and the distance sqrt(d^2 - (s/2)^2), where s is the distance sqrt((x2-x1)^2+(y2-y1)^2) between P1 and P2: +-----+P2 | / \ | s / \ | + \d | /P4 \ \ |/ \ \ +-----________ \ \ P1 d -----+P3 This gives you: x1+x2 sqrt(d^2 - (((x2-x1)^2+(y2-y1)^2)/4) * (y2 - (y1+y2)/2) x3 = ----- +/- ------------------------------------------------------- 2 sqrt((x2 - (x1+x2)/2)^2 + (y2 - (y1+y2)/2)^2) which I will let you simplify. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/