# Perpendicular Bisector?

• Jul 8th 2010, 05:43 AM
Punch
Perpendicular Bisector?
http://i952.photobucket.com/albums/a...g/P7080022.jpg
http://i952.photobucket.com/albums/a...g/P7080023.jpg

I have no idea how to start this question... I learnt how to do this a few years ago...
A few clues i have, one, use a compass, two, it's something called perpendicular bisector..
any help is appreciated, thanks in advance!!
• Jul 8th 2010, 07:58 AM
earthboy
the locus of any point P which is equidistant from two intersecting lines is the angular bisector of the included angle between the two lines .
here,as P is equidistant from AB and BC,draw the bisector of angle ABC.name the line BD.now the point P lies somewhere on line BD.
Now P is also equidistant from from A and B.so,draw the perpendicular bisector of AB.extend the bisector enough to intersect BD.thus,the point where the perpendicular bisector of AB meets angle bisector BD,is the exact location of point P.
its because :that will form two triangles PIA and PIB ,which will be congruent by side-angle-side congruency,and thus PA=PB(Happy)
hope that it is correct and helps!!!
• Jul 10th 2010, 04:15 AM
Punch
bring this up for more solutions, still stucked :( thx though
• Jul 10th 2010, 05:03 AM
bjhopper
draw any angle. intersection of rays is B. Legs are AB and CD of indefinite lenght. A ship starting at B must travel along the perpendicular bisector of ABC. You construct this line with your compass and straigtedge.If you do not know how google it.The ship moves along this line.Its distance from B depends on how far it travels. Given this distance the perpendicular distance to each leg can be calculated or determined by graphing

bjh