Three Points Equidistant...

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September 19th 2010, 05:05 PM
ragnar

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.
September 19th 2010, 05:20 PM
skeeter

Quote:

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.
September 19th 2010, 07:27 PM
Soroban

Hello, ragnar!

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

Quote:

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

Code:

| (-2,2) | (3,2) o - - - | - - - - - - - o P * | | * | | - - - - - - - |*- - - - - - - | - - - | ♥ | | * | | * | | * | | * Q | (3,-3)

$\,PQR$ is a right triangle.

Therefore, the circumcenter is the midpoint of the hypotenuse:

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