# Math Questions -- Algebra

Discussion in 'Physics & Math' started by Amadeusboy, Apr 4, 2008.

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Temur,
Thanks for taking the time to respond. Knowing positively that $p$ must be a constant would have simplified the question. I should have recognized this from the start. Oh well, live and learn.

Thanks again,

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I have a quick question. What would be the graph of

$\mid y+2\mid = \mid x-3\mid$?

Is is an "X" centered at $(-2,3)$ and having a 45 degree angle with $y=-2$?

Thanks in advance for any help.

5. ### temurman of no wordsRegistered Senior Member

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Yes. I think there is a typo: it is X centered at (3,-2).

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Thanks, Temur. You're right; there is a typo. It should be centered at $(3,-2)$.

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I came across another couple questions with which I had problems. Again, they are taken from p. 50 of Boyce and DiPrima Calculus (1988). Note that I added the bracketed definitions.

First,
Find an equation satisfied by all points twice as far from $(-1,3)[=A]$ as from $(4,2)[=B]$.​

and secondly,

Find an equation satisfied by all points equidistant from the origin and the line $x=-2$.​

In the first problem, I think my difficulty lies in the interpretation . Clearly, the equation will be some sort of circle, but is it centered on one of the two points mentioned or should one randomly select a point $P=(x,y)$ and find $x$ and $y$ such that $d(P,A)=2d(P,B)$. (I haven't done the algebra, but this will be some sort of circle.)

For the second problem, the graph of the equation in question must lie to the right of the line $x=-2$. Thus, any point satisfying this equation will have a distance of $x+2$ from the line $x=-2$ and a distance $\sqrt{x^2+y^2}$ from the origin. Would setting these two distances equal to each other and solving give the desired result? Is $x=y^2/4+-1$ the correct answer?

Thanks in advance for any assistance.

9. ### temurman of no wordsRegistered Senior Member

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1,330
The latter version is correct. Finding an equation for a curve means finding a relation between x and y where (x,y) is a point on the curve. (I doubt it is a circle though)

Yes, correct.

10. ### D HSome other guyValued Senior Member

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2,257
I suggest that you work with the square of the distances rather than distances because that way you won't have those nasty radicals to deal with. With this, your equation becomes $d^2(P,A)=4d^2(P,B)$ or $4d^2(P,B) - d^2(P,A) = 0$, where $d^2(P,A) = (x+1)^2+(y-3)^2$ and similar for $d^2(P,B)$. Note that $d^2[tex] is a positive definite form. In other words, you do not have to worry about a negative values for [tex]d^2$ (and worrying about them well get you in trouble, like it did on the next problem.)

No. That +- doesn't belong. Look at it this way. Set y to 0. There is only one point on the x-axis that is equidistant from the origin and the line x=-2: The point (-1,0). Your equation says (1,0) is also a solution. You got in trouble by worrying about radicals having two solutions when you should not have done so.

The answer to the first problem is a circle and the second, a parabola. In fact, the second answer is the standard geometric definition of a parabola as the locus of all points that are equidistant between a line and a point. The line is called the directrix of the parabola and the point, the focus of the parabola.