Question 43945
find an equation of the perpendicular bisector of line segment AB: A(-9,-3),
B(1,-7)
WE CAN USE THE PROPERTY OF PERPENDICULAR BISECTOR(LM SAY) TO SOLVE EASILY.ANY POINT ON IT IS EQUIDISTANT FROM A AND B.IF P (X,Y)IS ANY POINT ON IT,THEN
PA=PB
PA^2=PB^2
(X+9)^2+(Y+3)^2=(X-1)^2+(Y+7)^2
X^2+18X+81+Y^2+6Y+9=X^2-2X+1+Y^2+14Y+49
20X-8Y+40=0 IS THE EQN. OF PERPENDICULAR BISECTOR.