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Mugwump101
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Loci Mathemathics Problem
« on: Nov 22nd, 2005, 11:21pm » |
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I kind of need help with this one.  Explain it to me step by step and see if half my work is okay.
Problem:
36)
a) Write an equation of locus of points 5 units from (4,0)
b) Describe fully the locus of points d units from x-axis.
c) Find the # of points that simultaneously satisfy a) + b) if
(1) d= 2 (2) d=5 (3) d=10
d) Find the coordinates for all pts 5 units from (4,0) and 5 units from the x-axis.
I did try it and my results are incomplete but here's what I go:
First, I plotted the point and I was able to make a circle out of it by moving 5 units to all sides.
a) 25= (square root symbol) (x-4)^2 + (y-0)^2 ?
b) not sure what it means by describe and what I am supposed to be describing?
c) er.....
d) Coordinates: (4,5) ;(-1,-1); (9,0); (4,-5) ?
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towr
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Re: Loci Mathemathics Problem
« Reply #1 on: Nov 23rd, 2005, 4:59am » |
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on Nov 22nd, 2005, 11:21pm, Mugwump101 wrote:| a) 25= (square root symbol) (x-4)^2 + (y-0)^2 ? |
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Without the (square root symbol) this would be correct.
Otherwise you also need to take the square root of 25.
Always test your equation, try (9,0) for instance. It should be on the circle.
Quote:| b) not sure what it means by describe and what I am supposed to be describing? |
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You need to give an equation that defines all points at distance d from the x-axis.
hint, the x-axis has the equation y=0
Quote:Here you need to look at how many points satisfy both equations at the same time. Or in other words at how many points does the circle cross the lines parallel to the x-axis.
(part hidden because it gives away b)
Quote:| d) Coordinates: (4,5) ;(-1,-1); (9,0); (4,-5) ? |
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(9,0) lies on the x-axis, so it isn't 5 units away from it. And (-1,-1) is neither 5 units away from (4,0), or 5 units away from the x-axis. you probably meant (-1,0), but that's on the x-axis again)
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Icarus
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Re: Loci Mathemathics Problem
« Reply #2 on: Nov 23rd, 2005, 9:26pm » |
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To expand on towr's replies:
a) Remember that the formula for distance d between points (x,y) and (a,b) is
d2 = (x-a)2 + (y-b)2
or you can take the square root of both sides to get a direct equation for distance:
d = sqrt( (x-a)2 + (y-b)2)
Your problem is that you have mixed the two forms together, squaring d, and at the same time taking the square root of the other side.
b) You can describe this by one equation (if you set it up right), but two equations, or simply verbally with geometric descriptions.
On way to help you understand (without simply giving the whole thing away - you are supposed to be learning this yourself after all) is to ask: given an arbitrary point (x, y), how far is this point from the x-axis? I.e, How far is it from (2,5) to the line y=0? How about (2, -5)? What about other values of x and y? If you dont see it, try plotting the points on a graph and looking at how far they are from the x-axis. It shouldn't take you too long to recognize the pattern.
(Warning: the locus of points actually consists of two pieces - make sure you get both.)
c) If you plot the solution to (b) for each of the distances listed (on separate plots for each distance) and draw the circle on each plot. You can simply count the number of intersections for each distance. I would suggest that you do this instead of trying to figure it out analytically. It will help you develop some geometric intuition, which will help you to understand future problems.
(d) I don't have much to add to what towr said, except to suggest plotting the points again, and considering both conditions they must satisfy. If you get (b) and (c), then (d) should be obvious.
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| « Last Edit: Nov 23rd, 2005, 9:29pm by Icarus » |
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Mugwump101
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Re: Loci Mathemathics Problem
« Reply #3 on: Nov 24th, 2005, 8:47am » |
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Thank you, I'm understanding it A LOT more now. Graphing it was easier to see it and here's what I've got overall. Please check it. Thanks.....
a)25= (x-4)^2+ (y-0)^2
b) 5 units away from the radius
c)
(1) none
(2) 2 points [ (0,2) + (0, -2) ]
(3) 2 points [ 0, -8 ) (0, 8 ) ]
d) X-axis = (4,-5) and (4,5)
Other points are: (-1,0); (4,5); (9,0); (4,-5)
I graphed a circle with a radius of 5, then for c) I graphhed a circle with a radius of 2 and then another for a radius of 10. The points that fell on the y-axis are the points I listed for c).
Let me know if I'm not understanding something or I ignorantly made another mistake.
Thank you so much for your time in advance! 
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| « Last Edit: Nov 24th, 2005, 8:49am by Mugwump101 » |
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towr
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Re: Loci Mathemathics Problem
« Reply #4 on: Nov 24th, 2005, 9:14am » |
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on Nov 24th, 2005, 8:47am, Mugwump101 wrote:| I graphed a circle with a radius of 5, then for c) I graphhed a circle with a radius of 2 and then another for a radius of 10. The points that fell on the y-axis are the points I listed for c). |
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d is the distance to the x-axis, so you shouldn't change the radius of the circle; the circle stays the same, with a radius of 5.
So that leads to problems for c)
And you also shouldn't look at the points where the circle crosses the y-axis, but where it crosses the lines describing the points at a distance of d from the x-axis
c1) y=+/- 2
c2) y=+/- 5
c3) y=+/-10
And hopefully this will make you reconsider b) as well. 
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Icarus
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Re: Loci Mathemathics Problem
« Reply #5 on: Nov 24th, 2005, 8:33pm » |
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Indeed. I am not even sure what you could possibly mean by "5 units from the radius".
Perhaps the problem here is that you are not understanding what it means to be d units from a line? Draw a line and a point A not on it.
. A
___________________________________
For each point on the line, you could draw the line segment connecting the point to A and measure its length. However, of all these theoretical line segments, one is shorter than all the others - the one that is perpendicular to the line. This shortest distance between A and points on the line is what we mean the distance from A to the line.
More generally, given two figures (loci, sets, etc), the distance between them is defined to be the smallest distance between any point of one and any point of the other.
Now consider again our point and line. Can you plot other points that have the same perpendicular distance to the line as A does? Plot as many of them as it takes to recognize the pattern. And remember what I said earlier about the locus having two pieces. If you only come up with one, you should reflect to where the other would be.
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Mugwump101
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Re: Loci Mathemathics Problem
« Reply #6 on: Nov 27th, 2005, 10:38am » |
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Okay, would this be it:
I understand what you did for c in the sense that it moves positively and negatively the given units from the x-axis in result would equal thoses numbers. But for B............?
a) 25= (x-4)^2 + (y-0)^2
b) Move 5 units from the x-axis
c)
c1) y=+/- 2
c2) y=+/- 5
c3) y=+/-10
d)coordinates from all pts 5 units from (4,0) and 5 units from x-axis : (4,5) ; (4,-5)
I'm still not sure if what I'm doing is positively right but I think it should be. May you check it now? I'm sorry about having a hard time understanding it.
Thank you so much for your time again!!
For Icarus, the points are equidistant from each other in terms of units. In this chance, Point A is one unit away from the line.
. A .A
__________________________________
.A . A
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Re: Loci Mathemathics Problem
« Reply #7 on: Nov 27th, 2005, 11:03am » |
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a and d are correct.
But b and c are still a bit of a problem.
So
Quote:
b) Describe fully the locus of points d units from x-axis. |
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So we take a look at a random point (x1,y1), and it has to be d units from the x-axis
The equation for the x-axis is y=0, every point (xi, 0) lies on the x-axis, regardless of what xi is.
We want (x1,y1) to be d units from the x axis, it will be closest to the point on the x-axis which has the same x-coordinate (thus for xi=x1), so it will be closest to the point (x1,0)
the difference between (x1,y1) and (x1,0) is |y1|. And we want this distance to be d, so y1 = +/- d (plus d or minus d)
From this we know that any point (x1, +/- d) will be d units from the x-axis, regardless of the value of x1.
Therefor the equation(s) for the lines on which all these points lie is y=+/-d
For c), d get's a specific value, d=2, d=5 and d=10
But you still need to do something with it. Namely find how many times these lines (for a given distance d) cross the circle.
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| « Last Edit: Nov 27th, 2005, 11:13am by towr » |
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Mugwump101
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Re: Loci Mathemathics Problem
« Reply #8 on: Nov 27th, 2005, 11:51am » |
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okay, I think I'm understanding it better now...
C)
So, I add/subtract d to get y but what coordinate do I use.... Should I use (4,0)?
See I saw a pattern included with the coordinates (4,-5) where you subtract -5 from 4 and add 5 to -5 and you get (-1,0) and for (4,5) where you add 5 for 4 and subtract 5 from 5, you get (9,0) .
Do you mean I use that for coordinate (4,0) for c)
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Re: Loci Mathemathics Problem
« Reply #9 on: Nov 27th, 2005, 11:52am » |
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on Nov 27th, 2005, 10:38am, Mugwump101 wrote:| I'm sorry about having a hard time understanding it. |
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No problem. If either of us were able to actually have a regular conversation with you, we could of had this straightened out in a half-hour at most. But the restrictions of communicating through this forum (particularly with us posting at different times of the day) make it much harder.
But, as towr has done, it is time to stop trying to suggest the answer in hopes you will follow, but rather to be specific.
Your problem with (b) is that you did not do what the problem asked. It asked you to describe a locus of points. "Move 5 units from the x-axis" does not describe a locus! The whole point of the question is: "What collection of points do you get when you move 5 units away from the x-axis?"
The answer, as towr explains, is 2 horizontal lines, one 5 units above the axis, the other 5 units below:
________________________________
________________________________>
________________________________
All the points on these two lines are each 5 units away from the x-axis, and any point that is 5 units away from the x-axis is on one of these two lines. THAT is what it means to describe a locus!
And now, do you see why this is helpful? Consider problem (c). Graphically, it is easy to solve, once you know how to plot the set of all points at distance d from the x-axis:
For each value of d (do a separate graph for each value, otherwise you will have trouble with confusing which lines are for which value of d), graph the circle, and graph the two lines that represent all points d units from the axis. How many intersections are there?
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Re: Loci Mathemathics Problem
« Reply #10 on: Nov 27th, 2005, 12:02pm » |
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I see you replied to towr while I was creating my post.
You do not choose a point yourself. You have to find the point from solving the equations, if you needed to know its coordinates.
But (c) does not ask for the coordinates of the points. It simply asks for how many points there are. And this you can find out by some careful plotting. Do this on separate graphs for each value of d. I will describe it for d=2.
Plot your circle.
Plot the line y=2 (that is, y=d)
Plot the line y=-2 (that is, y=-d)
Count the intersections of these two lines with the circle. That is the number they are asking for.
The case d=5 can be a bit tricky, if your plot is less than perfect, so take great care there.
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