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# If equation |x/2|+|y/2| = 5 encloses a certain region

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If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  15 Jan 2012, 08:42
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If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A. 20
B. 50
C. 100
D. 200
E. 400
[Reveal] Spoiler: OA
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Re: Area of region [#permalink]  15 Jan 2012, 10:07
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Apex231 wrote:
I had solved till this point - So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

But instead of joining these points i did this - 4 * (10 * 10) = 400 , which is wrong of course.

So when we join these points, how |x|+|y| = 10 stays satisfied , what's the maths behind it?

Given: $$|x|+|y|=20$$

You will have 4 case:

$$x<0$$ and $$y<0$$ --> $$-x-y=10$$ --> $$y=-10-x$$;

$$x<0$$ and $$y\geq{0}$$ --> $$-x+y=10$$ --> $$y=10+x$$;

$$x\geq{0}$$ and $$y<0$$ --> $$x-y=10$$ --> $$y=x-10$$;

$$x\geq{0}$$ and $$y\geq{0}$$ --> $$x+y=10$$ --> $$y=10-x$$;

So we have equations of 4 lines. If you draw these four lines you'll get the figure shown in my previous post.

Hope it's clear.
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Re: Area of region [#permalink]  15 Jan 2012, 09:26
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Apex231 wrote:
Please move this to PS thread...

If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A 20
B 50
C 100
D 200
E 400

First of all to simplify the given expression a little bit let's multiply it be 2: $$|\frac{x}{2}|+|\frac{y}{2}|=5$$ --> $$|x|+|y|=10$$.

Now, find x and y intercepts of the region (x-intercept is a value(s) of x for y=0 and similarly y-intercept is a value(s) of y for x=0):
$$y=0$$ --> $$|x|=10$$ --> $$x=10$$ and $$x=-10$$;
$$x=0$$ --> $$|y|=10$$ --> $$y=10$$ and $$y=-10$$.

So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by $$|x|+|y|=10$$:
Attachment:

Enclosed region.gif [ 2.04 KiB | Viewed 7121 times ]
You can see that it's a square. Why a square? Because diagonals of the rectangle are equal (20 and 20), and also are perpendicular bisectors of each other (as they are on X and Y axis), so it must be a square. As this square has a diagonal equal to 20, so the $$Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200$$.

Or the $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

Answer: D.

Similar questions:
m06-5-absolute-value-108191.html
graphs-modulus-help-86549.html

Hope it's clear.
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Re: Area of region [#permalink]  15 Jan 2012, 09:58
I had solved till this point - So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

But instead of joining these points i did this - 4 * (10 * 10) = 400 , which is wrong of course.

So when we join these points, how |x|+|y| = 10 stays satisfied , what's the maths behind it?
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Re: Area of region [#permalink]  15 Jan 2012, 11:10
Thanks so much Bunuel...
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If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  10 Sep 2012, 11:25
CMcAboy wrote:
Can someone help me with this question:

If equation |x/2| + |y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A) 20
B) 50
C) 100
D) 200
E) 400

I believe this is the simplest & the quickest solution
|x/2| + |y/2| = 5
Put x = 0 in the above equation we get |y/2| = 5, which means y= 10, - 10
Put y = 0 in the above equation we get |y/2| = 5, which means x= 10, - 10

If you see plot these four points you get a square with two equal diagonals of length 20 units
Thus area = 1/2 * (Diagonal)^2 -----> 1/2 * 400 = 200

I hope this will help many.
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Re: If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  05 Dec 2012, 22:30
(1) derive all equations
x+y = 10
x-y = 10
x+y=-10
-x+y=10

(2) Get your x and y intercepts

(0,10), (10,0)
(0,-10),(10,0)
(0,-10),(-10,0)
(0,10),(-10,0)

(3) You will have a square with a diagonal of 20
(4) Calculate area = $$(10 * \sqrt{2})\sqrt{^2}$$ = 200

Answer : D
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Re: If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  12 Oct 2014, 02:26
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If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  15 Oct 2014, 18:17
Apex231 wrote:
If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A. 20
B. 50
C. 100
D. 200
E. 400

Hello There,
Equation of a straight line whose x and y intercepts are a and b resp. is (x/a) + (y/b) = 1 i.e., coordinates of two ends of the line are (a,0) and (0,b).
Now, from the given question,
|x/2|+|y/2| = 5, reducing this to intercept form we get,
|x/10|+|y/10| = 1
Considering the equation without modulus, coordinates are (10,0) and (0,10). Since there is modulus, other two coordinates are (-10,0) and (0,-10).
Now coordinates (10,0), (0,10), (-10,0) and (0,-10) form a square with diagonal length = 20.
Here diagonal length can be obtained by calculating the distance between (10,0) and (-10,0) or (0,10) and (0,-10).
In a square,
Diagonal = Side * sqrt(2)
Side = 10 * sqrt(2)
Area = Side * Side = 200.

Ans : D

Hope this helps!
Thanks!
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Re: If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  15 Oct 2014, 23:48
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Re: If equation |x/2|+|y/2| = 5 encloses a certain region   [#permalink] 15 Oct 2014, 23:48
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