Points $A,B,C$ are given, find a line and a point such that distance of $A,B,C$ from that line is equal to distance of $A,B,C$ from that point.

Question

In the name of God

Suppose points $A(1,2)$ and $B(2,5) , C(3,10)$ are given. Find equation of line $l$ and point $F$ such that $AH = AF , BH' = BF , CH'' = CF$, where $AH, BH' $and$ CH''$ are perpendicular lines to line $l$.

I first tried to solve this by solving system of equation. Suppose equation of line $l$ is equal to $y = mx+b$. and point $O$ is equal to $(x,y)$.

$m,b,x,y$ are unknown here, but we have only $3$ equalities (one for each point).
It is impossible to solve a system of equation with $4$ unknown variables and $3$ equalities.

I know the correct way of solving this question. There is only one parabola crossing $3$ points and we can find the equation of that parabola. By calculating the focus and the directrix of that parabola, we can find the answer.

But my question is why my method isn't working? How is it possible for a system of equation with $4$ unknown variables and $3$ equalities to have a unique answer?

Answer 1

There are infinite solutions to this problem. Each solution is a parabola.

To find these solutions, we parameterize them by the direction of the axis of the parabola.

Define a new coordinate frame $O x' y'$ such that

$ (x, y) = R (x' , y') $

where

$ R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $

We are given the points $A,B,C$ with coordinates defined in the $Oxy$ frame, from which we can derive their coordinates in the $Ox'y'$ as follows

$ A' = R^T A $

$B' = R^T B $

$C' = R^T C $

Now the parabola has the equation

$ y' = a x'^2 + b x' + c $

Substituting $A', B', C'$ gives a linear system of three equations in the unknowns $a,b,c$

Upon solving the system, we put the parabola in vertex format

$ y' = a (x' - x'_0)^2 + y'_0 $

where $x_0 = \dfrac{-b}{2a}$ , and $y'_0 = c - \dfrac{b^2}{4a} $

Define the parabola parameter $p = \dfrac{1}{4a}$

Then with respect to the frame $O'x'y'$ the focus of the parabola is

$ F' = (x'_0 , y'_0 + p ) $

And the directrix is

$ y' = y'_0 - p $

At the final step, we have to transform these back to the world frame, so that

$ F = R F' $

and

$ [0, 1] R^T (x, y) = y'_0 - p $

This equation can be re-written as follows

$ - \sin \theta \ x + \cos \theta y = y'_0 - p $

This completes the specification of the required point $F$ and the line $\ell$ as functions of the angle $\theta$.

Here are some parabolas with different orientation for the three given points in the problem.