1. ## Analytic geometry

1.) Find the coordinates of a point equidistant from (1,6) (5,-6) and (6,-1)

2.)Find the two points on the lines 2x+3y+4=0 which are not at a distance 2 from the line 3x+4y-6=0

Quiz tomorrow.. I really need to understand these.. Thanks!

2. Originally Posted by fayeorwhatsoever
1.) Find the coordinates of a point equidistant from A(1,6) B(5,-6) and C(6,-1)
All of these points are not collinear. Consider them as the vertices of a triangle ABC. Recall that the perpendicular bisector of the sides intersect at the circumcenter (which is the equidistant point you're looking for). By finding the intersection of at least two perpendicular bisectors passing through the midpoints between say, A and B and A and C, we find the equidistant point.

I'm not sure I understand 2.

3. 1). The set of all points equidistant from two fixed points is a line, the perpendicular bisector of the line segment with the fixed points as endpoints.
Using the distance formula, we have for two fixed points $(x_1,y_1)$ and $(x_2,y_2)$, the perpendicular bisector is:
$\sqrt{(x-x_1)^2+(y-y_1)^2}=\sqrt{(x-x_2)^2+(y-y_2)^2}$
$(x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)$
$x^2-2xx_1+x_1^2+y^2-2yy_1+y_1^2=x^2-2xx_2+x_2^2+y^2-2yy_2+y_2^2$
$-2xx_1+x_1^2-2yy_1+y_1^2=-2xx_2+x_2^2-2yy_2+y_2^2$
$2(x_2-x_1)x+2(y_2-y_1)y=(x_2^2-x_1^2)+(y_2^2-y_1^2)$

Pick two of the points (say (1,6) and (5,-6)) and compute their perpendicular bisector using the above formula. Then pick another pair from the points (say, (1,6) and (6,-1), for example), and compute that line. The point equidistant from all three points (1,6), (5,-6), (6,-1) must lie on both lines, and therefore is their intersection.

--Kevin C.