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Re: graphs_Modulus....Help [#permalink]
13 Dec 2012, 03:26
Expert's post
Bunuel wrote:
srini123 wrote:
Thanks Bunuel, I used similar method for a similar question and I got wrong answer the question was
what is the area bounded by graph\(|x/2| + |y/2| = 5\)?
I got hunderd since x=10 x=-10 y=10 y=-10
isnt the area 400 ? the answer given was 200, please explain
I think this one is different.
\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)
After solving you'll get equation of four lines:
\(y=-10-x\) \(y=10+x\) \(y=10-x\) \(y=x-10\)
These four lines will also make a square, BUT in this case the diagonal will be 20 so the \(Area=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\), area=200.
If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.
Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.
In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).
Hope it's clear.
Hii Bunuel. What is the best approach of finding the points of intersection in order to make the square. _________________
Re: graphs_Modulus....Help [#permalink]
13 Dec 2012, 03:30
Expert's post
Marcab wrote:
Bunuel wrote:
srini123 wrote:
Thanks Bunuel, I used similar method for a similar question and I got wrong answer the question was
what is the area bounded by graph\(|x/2| + |y/2| = 5\)?
I got hunderd since x=10 x=-10 y=10 y=-10
isnt the area 400 ? the answer given was 200, please explain
I think this one is different.
\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)
After solving you'll get equation of four lines:
\(y=-10-x\) \(y=10+x\) \(y=10-x\) \(y=x-10\)
These four lines will also make a square, BUT in this case the diagonal will be 20 so the \(Area=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\), area=200.
If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.
Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.
In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).
Hope it's clear.
Hii Bunuel. What is the best approach of finding the points of intersection in order to make the square.
I'd say substituting x=0 and y=0 in the equations of lines and making a drawing. _________________
Re: In the x-y plane the area of the region bounded by the [#permalink]
17 Jul 2013, 22:20
This is an interesting combo of absolute values, plotting lines in coordinate system and then finding the resulting figure's area. Thanks all for presenting the approach!
Re: graphs_Modulus....Help [#permalink]
21 Aug 2013, 20:51
Bunuel wrote:
srini123 wrote:
Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?
First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:
x=2 x=-2 y=2 y=-2
This lines will make a square with the side 4, hence area 4*4=16.
Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4.
The side of the square can't be 4, instead its sqrt(8) _________________
Re: graphs_Modulus....Help [#permalink]
22 Aug 2013, 02:21
Expert's post
honchos wrote:
Bunuel wrote:
srini123 wrote:
Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?
First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:
x=2 x=-2 y=2 y=-2
This lines will make a square with the side 4, hence area 4*4=16.
Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4.
The side of the square can't be 4, instead its sqrt(8)
The side of the square IS 4:
Attachment:
MSP39361d6ehgde6ie87a8800003827f7f92a367c60.gif [ 1.86 KiB | Viewed 999 times ]
What you mean by "sufficient"? |x+y| + |x-y| = 4 gives FOUR equations, as explained on page 1.
Hey Bunuel,
I think I confused some things here. Just this morning, I did Chapter 9 of MGMAT Strategy Guide 2. Here it's stated that, if I have an equation with 2 absolute value expressions and one variable, I only need to set up 2 cases. Case A, when the absolute values have the same sign, Case B when the absolute values have different signs. But since here we got 2 variables and two absolute value expressions, I think I have to set up 4 equations. I just confused what I read a bit...sorry!
If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.
In this case the area would be 8
The square is formed by lines x = 2, y = 2, x = -2 and y = -2 from the mod equation given in the question. You cannot turn it the way you did because you made the square smaller when you turned it. The side of the square needs to be 4 and diagonal needs to be 4*root2. Instead when you turned it, you made the diagonal 4 and side 4/root2 = 2*root2. That is incorrect. You can turn the square but it will not cut the axis at 2 or -2. It will cut the axis at 2*root2 or -2*root2. Then the square remains the same area-wise. _________________
Could you please elaborate more on this one . I am really having a hard time figuring this one out . I understood the first question i.e In x-y plane, the area of the region bounded by the graph of |x+y| + |x-y| = 4 . The area is 16. But in Second Question what is the area bounded by graph|x/2| + |y/2| = 5? Why was the graph drawn different from that of the previous Question. Why was diagonal considered for second question and Side of a square considered for first question.
Help is appreciated . Thanks in advance
VeritasPrepKarishma wrote:
pawankumargadiya wrote:
If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.
In this case the area would be 8
The square is formed by lines x = 2, y = 2, x = -2 and y = -2 from the mod equation given in the question. You cannot turn it the way you did because you made the square smaller when you turned it. The side of the square needs to be 4 and diagonal needs to be 4*root2. Instead when you turned it, you made the diagonal 4 and side 4/root2 = 2*root2. That is incorrect. You can turn the square but it will not cut the axis at 2 or -2. It will cut the axis at 2*root2 or -2*root2. Then the square remains the same area-wise.
Could you please elaborate more on this one . I am really having a hard time figuring this one out . I understood the first question i.e In x-y plane, the area of the region bounded by the graph of |x+y| + |x-y| = 4 . The area is 16. But in Second Question what is the area bounded by graph|x/2| + |y/2| = 5? Why was the graph drawn different from that of the previous Question. Why was diagonal considered for second question and Side of a square considered for first question.
Help is appreciated . Thanks in advance
The graph was drawn differently because the equations given to you are different.
|x/2| + |y/2| = 5 gives us a set of 4 equations. What are they?
y = 10+x y = 10 -x y = x - 10 y = -x - 10
When you draw them out, you get slanting lines and the figure which looks like a kite. The coordinates of the square will be (10, 0), (0, 10), (-10, 0) and (0, -10). Here, 20 is the length of the diagonal.
In our original question, when we draw out the 4 equations, we get horizontal/vertical lines and hence a regular looking square. In that case, 4 is the length of the side. _________________
I totally agree with your explanation, but the point is, why couldn't we draw the slant lines for the points (2,0), (-2,0), (0,2) and (0,-2) instead of horizontal lines and consider the length of diagonal rather than length of side for the original question (|x+y| + |x-y| = 4).
Thanks in advance.
VeritasPrepKarishma wrote:
dheeraj24 wrote:
Hi karishma,
The graph was drawn differently because the equations given to you are different.
|x/2| + |y/2| = 5 gives us a set of 4 equations. What are they?
y = 10+x y = 10 -x y = x - 10 y = -x - 10
When you draw them out, you get slanting lines and the figure which looks like a kite. The coordinates of the square will be (10, 0), (0, 10), (-10, 0) and (0, -10). Here, 20 is the length of the diagonal.
In our original question, when we draw out the 4 equations, we get horizontal/vertical lines and hence a regular looking square. In that case, 4 is the length of the side.
I totally agree with your explanation, but the point is, why couldn't we draw the slant lines for the points (2,0), (-2,0), (0,2) and (0,-2) instead of horizontal lines and consider the length of diagonal rather than length of side for the original question (|x+y| + |x-y| = 4).
Thanks in advance.
Because you are asked the area of the region bounded by |x+y| + |x-y| = 4. This equation gives you ONLY horizontal/vertical lines passing through points (2,0), (-2,0), (0,2) and (0,-2) such as x = 2, y = 2, x = -2, y = -2.
Note that x = 2 is the equation of a line (it is not a coordinate) which passes through point (2, 0) and is parallel to the y axis. Similarly, y = 2 is the equation of a line which is parallel to x axis and passes through the point (0, 2) and so on. I think you are taking x = 2 as a coordinate but that is not the case. A coordinate has a value for y too. x =2 is the equation of a line. It implies that x coordinate is always 2 and y can be anything. So all points lying on a line passing through x = 2 and parallel to y axis satisfy this criteria. _________________
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