# Thread: Equations of bisector + Intersection.

1. ## Equations of bisector + Intersection.

Hi all just wondered if anyone could tell me if my answers are right. They seem right but when I try to solve the simultaneous equation later they don't work.

So I had 3 points A(-7,4), B(-5,-2) and C(1,4)

To find the equation for the bisector of line AB I found the slope of the bisector to be 1/3, the midpoint to be (-6,1) and the equation of the bisector to be y=1/3x-3.

For the line BC I found the slope of the bisector to be -1, the midpoint to be (-2,1) and the equation of the bisector to be y=-1x+1.

Plotting them all the points on a graph they seem to work well. Only on trying to find the intersection of the bisector by having 1/3x-3=-1x+1 it all went a bit wrong. Estimating from the graph the point of intersection should be (-3,2). But I can't pull those answers out of the simultaneous equation.

So either I got the equations for the bisectors wrong, or I am doing the simultaneous equations wrong.

As always I really appreciate a nudge in the right direction.

2. ## Re: Equations of bisector + Intersection.

Originally Posted by alexpasty2013
Hi all just wondered if anyone could tell me if my answers are right. They seem right but when I try to solve the simultaneous equation later they don't work.

So I had 3 points A(-7,4), B(-5,-2) and C(1,4)

To find the equation for the bisector of line AB I found the slope of the bisector to be 1/3
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.

, the midpoint to be (-6,1) and the equation of the bisector to be y=1/3x-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.

For the line BC I found the slope of the bisector to be -1, the midpoint to be (-2,1) and the equation of the bisector to be y=-1x+1.
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.

Plotting them all the points on a graph they seem to work well. Only on trying to find the intersection of the bisector by having 1/3x-3=-1x+1 it all went a bit wrong. Estimating from the graph the point of intersection should be (-3,2). But I can't pull those answers out of the simultaneous equation.
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.

So either I got the equations for the bisectors wrong, or I am doing the simultaneous equations wrong.

As always I really appreciate a nudge in the right direction.

3. ## Re: Equations of bisector + Intersection.

Originally Posted by 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.
Thanks a lot that was really helpful post. Yeah I did mean the perpendicular bisector. I suspected I had the signs wrong, bit of a bad habit of mine I am trying to get out of. I think it is always helpful to see where you are go wrong. The mistakes tend to stay in your mind easier than the things you get right I keep finding!