How do u find the point of intersection of three perpendicular bisectors?

Answer 1

solve the equations of any two and make sure that the solution satisfies the equation of the third perpendicular bisector

Answer 2

If 3 perpendicular bisectors intersect, the point of intersection must be defined by the point of intersection of any 2 of them. Let the end points of the bisected lines be P1, P2, P3, P4, P6, and P6, with co-ordinates (a,b), (c,d), (e,f), (g,h), and (m,n), respectively.
P!P2 will have a midpont of (c-a)/2, (d-b)/2 and a slope of (d-b)/(c-a). Its bisector will have a slope of -(c-a)/(d-b) , and an equation of
y - (d-b)/2 = -(c-a)/(d-b)(x - (c-a)/2)
The P3P4 bisector has the equation
y - (h-f)/2 = -(g - e)/(h - f)(x - (g - e)/2)
Solving these equations for x and y yields the intersection of the bisectors:
+y - (d-b)/2 = -(c-a)/(d-b)(x - (c-a)/2)
-y + (h-f)/2 = +(g - e)/(h - f)(x - (g - e)/2)
(h-f)/2 - (d-b)/2 = +(g - e)/(h - f)(x - (g - e)/2) -(c-a)/(d-b)(x - (c-a)/2)
(h-f)^2/2 - (d - b)(h - f)/2 = (g - e)(x - (g - e)/2) -(c-a)(h - f)/(d-b)(x - (c-a)/2)
(d - b)(h-f)[(h - f) - (d - b)]/2 = (d - b)(g - e)x - (d - b)(g - e)(g - e)/2 -(c-a)(h - f)x + (c-a)(h - f)(c-a)/2
(d - b)(h-f)[(h - f) - (d - b)]/2 = [(d - b)(g - e) - (c-a)(h - f)]x - [(d - b)(g - e)(g - e) + (c-a)(h - f)(c-a)]/2
{(d - b)(h-f)(h - f - d + b) - [(d - b)(g - e)^2 + (h - f)(c-a)^2]/2} = [(d - b)(g - e) - (c-a)(h - f)]x

x = {(d - b)(h-f)(h - f - d + b) - [(d - b)(g - e)^2 + (h - f)(c-a)^2]/2}/[(d - b)(g - e) - (c-a)(h - f)]

y = -(c-a)/(d-b)({(d - b)(h-f)(h - f - d + b) - [(d - b)(g - e)^2 + (h - f)(c-a)^2]/2}/[(d - b)(g - e) - (c-a)(h - f)] - (c-a)/2) + (d-b)/2

I'm sure these rather formidable equations can be further simplified.

Answer 3

You have to solve two equation and find out common pointor intersection point. If this common point satisfy another equation then this point will be intersection point.