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A point $(x,y)$ moves so that its distance from the line $x=5$ is twice as great as its distance from the line $y=8$. Find an equation of the path of the point.

I got the two equations:

$$x-2y+11=0 \tag{1}$$

$$x+2y-21=0 \tag{2}$$

The book claims that a single equation for the path of the point would be the product of the two above equations. I don't understand why you would multiply them.

George V. Williams
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Jesse
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  • (In order to add a tag to the equation, use \tag{1}. I also changed your equations to be centered by using $$ instead of $.) – George V. Williams Jul 21 '13 at 19:36
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    Cool, thanks! Very helpful =) – Jesse Jul 21 '13 at 19:38
  • Imagine the same question but with respect to origin for simplicity. Then $x=2y$ and $x=-2y$ are parts of the answer because distance works with absolute value. If you multiply your equations then both are considered, as in $x^2-4y^2=0$. – Maesumi Jul 21 '13 at 19:39

2 Answers2

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Essentially, what you have is:

$$\Big(x-2y+11=0 \quad \text{OR}\quad x+2y-21=0\Big) \iff (x-2y+11)(x+2y-21) =0 $$

In words:

An ordered pair $(x, y)$ is a solution to either $\;(x - 2y + 11) = 0\;$ or else $\;(x + 2y - 21) = 0\;$

if and only if $\;(x, y)\;$ is a solution to $ (x-2y+11)(x+2y-21) = 0$.

amWhy
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  • What if an ordered pair solves both equations? Do we have an AND? What would we do with said equations, i.e. would we still multiply them? – Jesse Jul 21 '13 at 20:18
  • That might happen, say, in the event that the solutions have a common factor (in which case, that factor would need only be used once in the product of factors. $(y - x)(x+ y + 1) = 0$ and $(y - x)(2x - y - 3) = 0$ Both are solved when $y = x$, so you could write your solution as $(y - x)(x + y + 1)(2x - y - 3) = 0$. In any case, if any ordered pair may satisfies two equations - It would be represented in the final equation (the product of solutions). – amWhy Jul 21 '13 at 20:24
  • I see that when we have an OR statement (Our solutions are A OR B) we multiply the two equations to get all possible solutions. But what I am wondering- is there a scenario where "Our solutions are A AND B", if so what do we do with the solutions then? – Jesse Jul 21 '13 at 20:31
  • Also, would it be correct to have $(y-x)^n$ as a factor in your solution? wouldn't it still yield the same result? – Jesse Jul 21 '13 at 20:33
  • Recall, we want to convey all, and only the information conveyed by separate equations. When multiplying by $(x−y)^n$ instead of say, x−y, in the final equation, you will likely introduce solutions in the "product" equation that you didn't have with the equations you started with – amWhy Jul 21 '13 at 20:41
  • Wait, so in your above equations, as long as we write the solution with a $(y-x)$ factor, the final equation could look like the product of anything? – Jesse Jul 21 '13 at 20:43
  • I don't understand how solutions would be introduced, if we have x=y. – Jesse Jul 21 '13 at 20:44
  • No: we want to express only the solutions to the given equations. What you're proposing would leave out some solutions, and may add other solutions. – amWhy Jul 21 '13 at 20:45
  • let us continue this discussion in chat – Jesse Jul 21 '13 at 20:52
  • @Jesse: If you’re working in $\Bbb R$, the set of $x$ such that $A(x)=0$ and $B(x)=0$ is the set of solutions to the equation $A(x)^2+B(x)^2=0$. – Brian M. Scott Jul 21 '13 at 20:57
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The graph of the product of two equations (each equalling zero), is the union of the graphs of each equation separately. In this case, there are two valid paths that might work.

In one dimension, the solutions to $(x-1)(x+2)=0$, are the union of the solutions to $x-1=0$ and $x+2=0$. (namely, $x=1$ and $x=-2$).

vadim123
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