Equidistant point

Question

I have to find the point $P(a,b)$ equidistant from $A(6,8),B(2,4)$ and $C(0,2)$.

Since $$\begin{array}{l}d_{PA}^2=a^2-12a+b^2-16b+100 \\ d_{PB}^2=a^2-4a+b^2-8b+20 \\d_{PC}^2=a^2+b^2-4b+4 \end{array}$$ then from $d^{2}_{PA}=d^{2}_{PB}$ and $d^{2}_{PA}=d^{2}_{PC}$, I get $$\left\{\begin{array}{l}-8a-8b=-80 \\ -12a-12b=-96\end{array}\right.\Rightarrow\left\{\begin{array}{l}a+b=10\\a+b=8\end{array}\right.$$ So may I conclude there's $P$ which is equidistant from $A,B$ and $C$?

Answer 1

(This is mostly a summary of the comments)

No, you may not conclude there's $P$ equidistant from $A, B$, and $C$, as can be seen with your equations.

Another way to see it is to note that $PA = PB$ implies that $P$ is on the perpendicular bisector of $AB$, and similarly for $AC$, $BC$. However, since $A, B$, and $C$ are collinear, the three perpendicular bisectors are parallel, and thus there does not exist a point $P$

In general, however, as mentioned in the comments, there is one when $A, B$, and $C$ are not collinear, because then they form a triangle, which has a known point $P$, known as the circumcenter of triangle $ABC$.

Answer 2

Due to the three points being collinear, one cannot form a circle passing through all 3 points, so the equidistant point is either non-existent or belongs at the point of infinity.

Answer 3

All three points $A(6,8),B(2,4)$, and $C(0,2)$ satisfy the equation $$y=x+2$$

Thus they are collinear and finding a point on the plane equidistance to these points is impossible.

Answer 4

Since no one else mentioned it:

The conclusion of impossibility in the posted question is accurate. However, at the time of this answer, the math shown in the posted question is inaccurate.

The difference equations should be

$-8a -8b = 80 \implies (a + b) = -10$.

$-12a - 12b = 96 \implies (a + b) = -8.$

Addendum

The alternative approach to this type of problem is to find the equation of the perpendicular bisectors between (for example) points A and B, and between points B and C.

However, as has been pointed out, since all 3 points are collinear, the slope of the 2 perpendicular bisectors will both be the same. That is, both perpendicular bisectors will have slope $(-1)$.

This means that the two perpendicular bisectors will be parallel lines, which means that they will never intersect. Therefore, since there can be no intersection between the two perpendicular bisectors, there can be no extraneous point equidistant from the $3$ points.