3

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.

Howcan
  • 395
  • On rearrangement, $$(y-6)^2-(x-2)^2=6^2-2^2=32$$ which is a hyperbola which is strongly related to https://people.richland.edu/james/lecture/m116/conics/hypdef.html – lab bhattacharjee Feb 10 '14 at 07:20
  • Interesting, I'll read that soon, thank you! – Howcan Feb 10 '14 at 07:25
  • Your quadratic equation has infinitely many solutions, you need an additional condition to determine $x$ and $y$. There are infinitely many pairs of points on the two axes equidistant from $(2,6)$. – André Nicolas Feb 10 '14 at 07:26
  • Hmmm, that's an interesting point. Well the condition was that the segment between the y axis and (2,6) was equal to the segment between the x axis and (2,6). – Howcan Feb 10 '14 at 07:28
  • Plus the condition that the line joining the two points actually passes through $(2,6)$. – André Nicolas Feb 10 '14 at 07:33
  • The "additional condition" is the linear equation you gave, $ \ y \ = \ 12 - 3x \ . $ Upon insertion of this into the equation of the hyperbola, that equation reduces to $ \ 8x^2 - 32x \ = \ 0 \ , $ which has the two solutions $ \ x = 0 \ \ \text{and} \ \ x = 4 \ . $ – colormegone Feb 10 '14 at 07:36

3 Answers3

2

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.

André Nicolas
  • 507,029
1

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.

enter image description here

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".]

colormegone
  • 10,842
  • This is very nice I appreciate the effort, definitely clears things up. So after I get the equation of the hyperbola, I could find the x and y intercepts and use the ones that go through 2,6 (I think that when you solve there aren't that many solutions, 2 y-ints and 2 x-ints from looking at the graph).

    It also turns out that we'll be doing conic sections soon, thanks.

     – Howcan Feb 10 '14 at 07:49
  • It is indeed the case that the hyperbola has two $ \ x-$ and two $ \ y-$ intercepts. However, only one line links a pair of them at once. In this particular situation, the other $ \ x-$ intercept is also a $ \ y-$ intercept (the origin!) and the line passing through it and $ \ (2,6) \ $ does not meet the hyperbola at an intercept. So the solution found is unique. – colormegone Feb 10 '14 at 07:54
  • What program did you use to make this graph? – Howcan Feb 10 '14 at 19:17
  • Just the Grapher utility on my Mac. – colormegone Feb 10 '14 at 20:04
0

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: $$ 2k+m = 6 \Rightarrow 2\cdot-3+12=-6+12=6. \qquad\blacksquare $$

Tommy L
  • 329
  • 2
  • 7
  • Actually, the answer is also in the quadratic equation you gave. Set $y=0$ and you have $x^2=4x$, which gives $x=0$ or $x = 4$. – Tommy L Feb 10 '14 at 07:52