Find the geometric locus of midpoints of segments connecting a given point with points lying on a given plane.



  • Point $P$
  • Plane $M$


Mark an arbitrary point $Q$ on $M$.

Draw line segment $l$; $l$=$PQ$.

Bisect $l$ at some resultant point $R$.

Draw plane $N$; $N || M$, $R$ lies on $N$.


$N$ is the geometric locus of midpoints of segments connecting a given point with points lying on a given plane.


I can give you the answer, as well as the construction, however I don't know how to prove that my solution is legit.


  • This exercise is found in Kiselev's Geometry; Book II: Stereometry (English adaptation).
    • It is Exercise 17, in Chapter 1: Lines and planes, Section 2: Parallel lines and planes.
  • $\begingroup$ "I'm having a hard time proving it!" Proving what? The question asks you to find the locus. Do you have a guess as to what the locus should be? $\endgroup$ – Joey Zou Sep 2 '16 at 5:00
  • $\begingroup$ @JoeyZou : Yes, I do know it, and even how to construct it. I've updated my question. However, how can I prove that my answer is indeed correct? $\endgroup$ – Fine Man Sep 2 '16 at 5:02
  • $\begingroup$ There is a basic flaw in your proof: It is true you have obtained the right plane, by using a particular point. How can you be sure that by taking another point, and applying the same process, you will find the same plane ? $\endgroup$ – Jean Marie Sep 2 '16 at 8:45
  • $\begingroup$ @JeanMarie : That's exactly my dilemma. Intuitively I know I'm right, but how can I prove that regardless of what point $Q$ I take, I'll get the same result? $\endgroup$ – Fine Man Sep 2 '16 at 17:21

Your transformation is an homothetic transformation $h$ whose center is the given point $S$ and ratio $\lambda=\frac{1}{2}$.

The image of the plane by $h$ is another plane. More precisely the image is the plane passing through the images of three independent points lying on the initial plane.


I complement the (very accurate) solution of @ mathcounterexamples.net by an analytical proof:

If the plane $(P)$ has equation


and the point, say $P_0$, has coordinates $(x_0,y_0,z_0)$, the locus of midpoints is the set of points having coordinates $$\begin{cases}X&=&\frac{x+x_0}{2}\\Y&=&\frac{y+y_0}{2}\\Z&=&\frac{z+z_0}{2}\end{cases} \ \ \ \Leftrightarrow \ \ \ \begin{cases}x&=&2X-x_0\\y&=&2Y-y_0\\z&=&2Z-z_0\end{cases}$$

Plugging these expressions into (1), we get:


which induces the following constraint on the set of points $(X,Y,Z)$:

$$\tag{2} uX+vY+wZ=\dfrac{1}{2}(ux_0+vy_0+wz_0)$$

The geometrical interpretation of (2) is that the set of points $(X,Y,Z)$ belongs to a certain plane $(P')$ parallel to the initial plane $(P)$ (they share the same normal vector $(u,v,w)$).

Moreover, as we have proceeded by equivalent conditions, the solution is the whole plane.

  • $\begingroup$ This isn't what I'm looking for, but I won't vote you down. I want the answer to be just Euclidean-geometric, not algebraic-geometric. The book I go by isn't algebraic-geometry. Hence, I didn't add "algebraic-geometry" tag. $\endgroup$ – Fine Man Sep 2 '16 at 18:32

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