Find an equation of the line cut off by the axes such that the midpoint is (2,6) I'm currently taking precalc level math and I was given a question, which I solved, I was just wondering about an alternate way to do it. 
"Find an equation of the line that passes through the point (2, 6)
in such a way that the line cut off between the axes is bisected by
the point (2, 6)"

It was fairly easy if you set (2,6) as the midpoint and solve for the x and y-intercepts, from which you can find the slope of -3, which gives the equation y=-3x+12. After that I decided to try to set 2 distance formulas equal to each other, one from the point (2,6) to (0, y), and the other from the point (2,6) to (x, 0) and began solving. You end up with 
x^2-4x = y^2 - 12y
and of course I have no idea how to solve that. So, how would you solve this and what topic/level of math would such an equation be? (I put the last equation into wolfram alpha and one of the answers did give the correct x and y intercepts)
I know this is really counter-intuitive but I was curious.
 A: Draw the line through $(2,6)$ parallel to the $x$-axis. Let it meet the $y$-axis at $A$, and let the $y$-intercept be $B$. 
Since $(2,6)$ bisects the hypotenuse, we have $OB=(2)(OA)$ and therefore the $y$-intercept is $12$.
Similarly, the $x$-intercept is $4$.
Now the equation of our line is easy to find. 
Remark: If you really want to use distances, note that by geometry the circle with centre $(2,6)$ and the intercepts passes through the origin. 
Thus 
$$(x-2)^2+(y-6)^2=2^2+6^2.$$
For the $y$-intercept, put $x=0$. We get $(y-6)^2=6^2$, and therefore $y=12$. The $x$-intercept is found in the same way. 
A: Here is the hyperbola which lab bhattacharjee produces from your distance equation; the two branches indicate where all the possible pairs of points lie for which $ \ (2,6) \ $ is the midpoint of the connecting segment.  The "constraining condition" which  André Nicolas mentions (your linear equation) picks out the specific segment marked in red.

This is still within the scope of pre-calculus (analytic geometry and conic sections), but may be a type of problem you haven't seen worked out yet.
[I put in this answer mainly to show the graph, so there's no need to give this "points".]
A: You can find the solution as follows:
State what you know from the given point and the intersection with the axes:


*

*$2\cdot k+m=6$

*$x\cdot k+m=0$

*$0\cdot k+m=y$


Express $m$ as a function of $k$:
$$
    m = 6 - 2k.
$$
Use the two last equations to solve for $k$ by inputting the first equation:
$$
    xk+6-2k=0,
$$
solve for k:
$$
    k=\frac{-6}{x-2}
$$
Do the same for $y$:
$$
    k=\frac{y+6}{-2}
$$
Set those expressions for $k$ equal and input $y=0$:
$$
    \frac{-6}{x-2} = \frac{y+6}{-2}
$$
and solve for x:
$$
    y + 6 = \frac{12}{x-2} \Rightarrow 6x-12=12 \Rightarrow x=4.
$$
Thus, at $x=4$, $y=0$ for the $k$ we are looking for.
Now you have two points on the line and thus the line is completely defined.
Solve for $k$ when $y=0$:
$$
    k = \frac{-6}{x-2} = \frac{-6}{4-2} = \frac{-6}{2} = -3.
$$
Solve for $m$:
$$
    m = 6 - 2k = 6 - 2\cdot-3 = 12.
$$
Thus, your line has equation $y = -3x+12$, and the other intersection point is at $y=12$.
Check the answer:
$$
  2k+m = 6 \Rightarrow 2\cdot-3+12=-6+12=6. \qquad\blacksquare
$$
