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**HallsofIvy** A "bisector" is any point or line (or other object) that divides the line segment into equal parts. If you mean the **perpendicular** bisector, then, yes, the slope of AB is (4-(-2))/(-7-(-5))= -6/2= -3 so the slope of the perpendicular is 1/3.

Yes, ((-7-5)/2, (4- 2)/2)= (-6, 1). But did you not notice that y= (1/3)(-6)- 3= -2- 3= -5, not 1? (If you meant y= (1/3)(x- 3) then y= (1/3)(-6- 3)= (1/3)(-9)= -3, also incorrect.) The "perpendicular bisector" of AB has equation y= (1/3)(x+ 6)+ 1= (1/3)x+ 3.

again, you are talking about the **perpendicular** bisector, right? Yes, the slope of BC is 1 so the slope of the perpendicular bisector is -1. Yes, the midpoint of BC is (-2, 1) but y= -1(-2)+ 1= 3, not 1. The equation of the perpendicular bisector of BC is y= -1(x- (-2))+ 1= -x- 1. You have the sign on the constant terms wrong on both equations.

I suspect you "plotted" the lines using the true points of bisection rather than plotting the equations you have here.

Solving y= (1/3)x+ 3 and y= -x- 1, we have (1/3)x+ 3= -x- 1 so that (1+ 1/3)x= -3-1 or (4/3)x= -4, x= -3. Putting that into y= (1/3)x+ 3 gives y= (1/3)(-3)+ 3= -1+ 3= 2. Similarly, putting x= -3 into y= -x- 1 gives y= -(-3)- 1= 3- 1= 2.