Perpendicular Bisector

Hey guys, something that is probably easy for you is lurking below:

We're working on Perpendicular bisectors at the minute, and all my working is perfectly right, but I still seem to be getting the wrong answer. I need to get a hang of it by tomorrow morning (It's currently 10pm) as we are moving on and won't be covering it again.

The exercises which we are using are:

1) Find the equation of the perpendicular bisector of the line joining each pair of points:

A(5, -7) and J (6,4)

I know that I need to get to y-b = m(x-a) and to do this I must first need to figure out the middle point using:

x1+x2 / 2

y1+y2 / 2

Which will fill in a and b of y-b = m(x-a)

I then need to use:

m = y2-y1 / x2-x1


then using the answer to this, I then need to use m1 x m2 = -1 to be able to fill in M in the y-b = m(x-a).

I then need to switch it around so that x + y + n = 0
(where n is number)

But I still seem to be getting the wrong answer. If someone could show me the solution to the above one, I might be able to figure out where I am going wrong.

Also, I have no idea where to start with:

Find the equation of the perpendicular bisector of each side of a triangle with vertices:

P (1, 3), R (0, 4) and Q (5, 2)

Any help would be much appreciated. I think that once I get a grasp of where I am going wrong with the first part, I should be able to figure out the second question.

Thank you so much in advance!

Quote:

---

Originally Posted by Mathy

1) Find the equation of the perpendicular bisector of the line joining each pair of points:

A(5, -7) and J (6,4)

---

Find the midpoint:




Now find the slope of the line connecting points A and J:


So the line we want has a slope of and intersects the point .

So




So




So your perpendicular bisector is


The graph below shows everything. (Sorry the bisector doesn't look perpendicular. The axes are scaled differently and I couldn't get it to "square up" nicely.)

-Dan