Question 146948
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 AB.
My answer is: y= -3x +3 
<pre><font size = 4 color = "indigo"><b>
Sorry, that's wrong. Plot the two points:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
locate(-2-.2,3+.35,"@"),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4.8-.2,3+.35,"A(-2,3)")
 
)}}}

Find the midpoint using the midpoint formula:

Given the two points ({{{x[1]}}}, {{{y[1]}}}), ({{{x[2]}}}, {{{y[2]}}}), 

Their midpoint = ({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

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

Their midpoint = ({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

= ({{{(-2+6)/2}}}, {{{(3+7)/2}}}) = ({{{4/2}}}, {{{10/2}}}) = ({{{2}}}, {{{5}}})

So we plot that, and connect the three points:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10), locate(-2-.2,3+.35,"@"), 
 locate(2-.2,5+.35,"@(2,5)"), line(-2,3,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4.8-.2,3+.35,"A(-2,3)")

)}}}


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

{{{m=(y[2]-y[1])/(x[2]-x[1]) = ((7)-(3))/((6)-(-2)) = 4/(6+2) = 4/8 = 1/2}}}

To find the slope of a line which is perpendicular to 
a line with slope {{{1/2}}}, we invert the fraction and 
change its sign, and get {{{-2/1}}} or {{{-2}}}

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

{{{y-y[1]=m(x-x[1])}}}

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

y=-2x+9

Now we draw that and get:

{{{drawing(370,400,-10,10,-10,10, graph(370,400,-10,10,-10,10,-2x+9), locate(-2-.2,3+.35,"@"), 
 locate(2-.2,5+.35,"@(2,5)"), line(-2,3,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4.8-.2,3+.35,"A(-2,3)")

)}}}

---------------------------------
</pre></font></b>

b.) Find an equation for the perpendicular bisector of BC.
y= -3x + 13
<pre><font size = 4 color = "indigo"><b>
Sorry, that's wrong, too Plot the two points:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
locate(-1-.2,6+.35,"@"),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4-.2,6+.35,"C(-1,6)")
 
)}}}

Find the midpoint using the midpoint formula:

Given the two points ({{{x[1]}}}, {{{y[1]}}}), ({{{x[2]}}}, {{{y[2]}}}), 

Their midpoint = ({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

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

Their midpoint = ({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

= ({{{(-1+6)/2}}}, {{{(7+6)/2}}}) = ({{{5/2}}}, {{{13/2}}}) = ({{{2.5}}}, {{{6.5}}})

So we plot that, and connect the three points:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10), locate(-1-.2,6+.35,"@"), locate(2.5-.2,6.5+.35,"@(2.5,6.5)"),
 locate(6-.2,7+.35,"@B(6,7)"), line(-1,6,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4-.2,6+.35,"C(-1,6)")

)}}}


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

{{{m=(y[2]-y[1])/(x[2]-x[1]) = ((6)-(7))/((-1)-(6)) = (-1)/(-7) = 1/7}}}

To find the slope of a line which is perpendicular to 
a line with slope {{{1/7}}}, we invert the fraction and 
change its sign, and get {{{-7/1}}} or {{{-7}}}

Now since the line goes through (2.5,7.5), we use the point-slope
form of a line's equation using {{{m=-7}}}:

{{{y-y[1]=m(x-x[1])}}}

{{{y-7.5=-7(x-2.5)}}}

{{{y-7.5=-7x+17.5}}}

{{{y = -7x + 24}}}

Now we draw that and get:

{{{drawing(370,400,-10,10,-10,10, graph(370,400,-10,10,-10,10,-7x+24), locate(-1-.2,6+.35,"@"), locate(2.5-.2,6.5+.35,"@(2.5,6.5)"),
 locate(6-.2,7+.35,"@B(6,7)"), line(-1,6,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4-.2,6+.35,"C(-1,6)")

)}}}



</pre></font></b>
c.) Find coordinates for a point K that is equidistant from A, B, and C.
<pre><font size = 4 color = "indigo"><b>

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. 

{{{drawing(370,400,-10,10,-10,10, graph(370,400,-10,10,-10,10,-2x+9), locate(-2-.2,3+.35,"@"), 
  line(-2,3,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4.8-.2,3+.35,"A(-2,3)"),
graph(370,400,-10,10,-10,10,-7x+24), locate(-1-.2,6+.35,"@"), 
 locate(6-.2,7+.35,"@B(6,7)"), line(-1,6,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4-.2,6+.35,"C(-1,6)")

)}}}

Now we can draw in the circle:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10,-2x+9), locate(-2-.2,3+.35,"@"), 
  line(-2,3,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4.8-.2,3+.35,"A(-2,3)"),
graph(400,400,-10,10,-10,10,-7x+24), locate(-1-.2,6+.35,"@"), 
 locate(6-.2,7+.35,"@B(6,7)"), line(-1,6,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4-.2,6+.35,"C(-1,6)"),
circle(3,3,5)

)}}}

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

{{{y = -2x + 9}}}
{{{y = -7x + 24}}}

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.

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10,-2x+9), locate(-2-.2,3+.35,"@"), 
  line(-2,3,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4.8-.2,3+.35,"A(-2,3)"),
graph(400,400,-10,10,-10,10,-7x+24), locate(-1-.2,6+.35,"@"), 
 locate(6-.2,7+.35,"@B(6,7)"), line(-1,6,6,7),
locate(6-.2,7+.35,"@B(6,7)"), locate(-4-.2,6+.35,"C(-1,6)"),
circle(3,3,5), locate(3-.2,3+.35,"@(3,3)") 

)}}}

Edwin</pre>