SOLUTION: This problem has 3 parts. I figured out the answer to a and b. i only need c. Let A = (-2,3), B = (6,7), and C = (-1,6). a.) Find an equation for the perpendicular bisector of

Sorry, that's wrong. Plot the two points:



Find the midpoint using the midpoint formula:

Given the two points (, ), (, ), 

Their midpoint = (, )

Substituting points (-2,3) and (6,7), 

Their midpoint = (, )

= (, ) = (, ) = (, )

So we plot that, and connect the three points:




Next we find the slope of AB using the slope formula:



To find the slope of a line which is perpendicular to 
a line with slope , we invert the fraction and 
change its sign, and get  or 

Now since the line goes through (2,5), we use the point-slope
form of a line's equation using :



y-5=-2(x-2)
 
y-5=-2x+4

y=-2x+9

Now we draw that and get:



---------------------------------

Sorry, that's wrong, too Plot the two points:



Find the midpoint using the midpoint formula:

Given the two points (, ), (, ), 

Their midpoint = (, )

Substituting points (6,7) and (-1,6), 

Their midpoint = (, )

= (, ) = (, ) = (, )

So we plot that, and connect the three points:




Next we find the slope of BC using the slope formula:



To find the slope of a line which is perpendicular to 
a line with slope , we invert the fraction and 
change its sign, and get  or 

Now since the line goes through (2.5,7.5), we use the point-slope
form of a line's equation using :









Now we draw that and get:

This amounts to finding the center of a circle that passes 
through all three points, for the center of a circle is
equidistant from all points on a circle.

AB and BC are both chords.  There is a theorem that says,

"The perpendicular bisectors of two chords intersect at the
center of a circle. 



Now we can draw in the circle:



So we solve the system of the equations of the two perpendicular 
bisectors of the above two problems and we get:




Solve that system of equations by substitution, which I assume
you can do, and get

x=3, y=3.

So the point (3,3) is the center of the circle, which is
equidistant from all three given points A, B, and C.



Edwin