1. ## Three Points Equidistant...

So I thought I knew how to solve this kind of question, but apparently not.

Find a point equidistant from the three points, P(3, 2), Q(3, -3), R(-2, 2).

My (wrong) answer: Mid-point of PQ is (3, -1/2), so the slope of the line from R to this point is (-5/2)/5 = -1/2. Going two to the right from R by this slope, we go down to 1, so the equation of the line is y = (-1/2)x + 1.

Mid-point of PR is (1/2, 2), so the slop of the line from this point to Q is -5/(5/2) = -2. Going from this point, one-half unit to the left we go up one, so the equation of the line is y = -2x + 3.

Now I want to find the point of intersection between these two lines, so I set them equal to each other and solve for x, giving me x = 4/3, which is wrong.

2. Originally Posted by ragnar
So I thought I knew how to solve this kind of question, but apparently not.

Find a point equidistant from the three points, P(3, 2), Q(3, -3), R(-2, 2).

My (wrong) answer: Mid-point of PQ is (3, -1/2), so the slope of the line from R to this point is (-5/2)/5 = -1/2. Going two to the right from R by this slope, we go down to 1, so the equation of the line is y = (-1/2)x + 1.

Mid-point of PR is (1/2, 2), so the slop of the line from this point to Q is -5/(5/2) = -2. Going from this point, one-half unit to the left we go up one, so the equation of the line is y = -2x + 3.

Now I want to find the point of intersection between these two lines, so I set them equal to each other and solve for x, giving me x = 4/3, which is wrong.
the point equidistant from all three points P, Q, and R is the intersection of the perpendicular bisectors of PQ, QR and PR

PQ is a vertical line ... the equation of the perpendicular bisector is y = -1/2

PR is a horzontal line ... the equation of the perpendicular bisector is x = 1/2

the point of intersection is (1/2, -1/2), which is equidistant from the three points.

fyi ... the point equidistant from the three vertices of a right triangle is always the midpoint of the hypotenuse ... make a sketch on graph paper and check it out.

3. Hello, ragnar!

If you had made a sketch, you would have seen skeeter's FYI immediately.

$\displaystyle \text{Find a point equidistant from the points: }\;P(3, 2),\,Q(3, \text{-}3),\; R(\text{-}2, 2).$
Code:
                |
(-2,2)     |             (3,2)
o - - - | - - - - - - - o P
*    |               |
* |               |
- - - - - - - |*- - - - - - - | - - -
|   ♥           |
|      *        |
|         *     |
|            *  |
|               * Q
|             (3,-3)

$\displaystyle \,PQR$ is a right triangle.

Therefore, the circumcenter is the midpoint of the hypotenuse:

. . . $\displaystyle \left(\frac{1}{2},\:-\frac{1}{2}\right)$