If equation |x| + |y| = 5 encloses a certain region on the graph, what

Question

If equation  |x| + |y| = 5  encloses a certain region on the graph, what is the area of that region?

A. 5
B. 10
C. 25
D. 50
E. 100

M06-05

Bunuel · Accepted Answer

X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0) (just make x equal to zero and find y and then make y equal to zero and find x). Now if we join these points we'll get the following region: https://gmatclub.com/forum/download/file.php?id=47214 The diagonals of the quadrilateral are equal (10 and 10), and also are perpendicular bisectors of each other (as they are on X and Y axis), so the figure must be a square. Area of a square equals to \frac{ ext{diagonal}^2}{2}=\frac{10^2}{2}=50 . Answer: D m06-05.gif

georgethomps · Answer

What about coordinate points (-2, 3), (2, 3), (-2, -3), (2, -3)? Those satisfy the equation and yield a different area

Area: (2 - (-2)) * (3 - (-3)) = 24

I don't understand why this isn't valid?

m1033512 · Answer

You may have combination of 1,4 also which will satisfy the given equation .

BUT THIS IS NOT CORRECT

Actually given equation is a collection of 4 straight lines which are caluclated on below logic

when x and y both negative
x negative and y positive 
x positive and y negative 
x positive and y positive

this way we get four lines amd then draw them on the xy plane

and calculate the enclosed area .

award kudos if helpful

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georgethomps · Answer

I'm still confused. Possible combinations of (5,0), (4,1), (3,2) all yield inconsistent results, wouldn't this be an invalid problem? What is wrong about using (1,4) or (2,3)? They both satisfy the equation and yield straight lines on the graph. I'm guessing this is supposed to be wrong because only (0,5) doesn't plot vertical lines, but I don't see anything in the problem that prevents this. The problem seems to state plotting the coordinates and interpreting them visually.

aliakberza · Answer

Note that the question has asked for the  area enclosed by the equation . In other words it is asking for the area where all these lines (4 in this case) intersect. From the equation we can see that the lines will intersect at (5,0), (0,5), (-5,0) and (0,-5).  While your coordinates may satisfy the equation, they are not the points at which these lines intersect . To make it simpler, when asked to compute area under the graph in questions of this type, look for the coordinates at which the lines intersect and compute the area enclosed by them.

Hope its clear now.

yg01 · Answer

Why can't the area be computed as side^2 where each side equals 5 so area would be 25

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Bunuel · Answer

You could find the area that way but notice that the side does not equal to 5, it actually equals to  5\sqrt{2}  (for example, the distance between two vertices (0, 5) and (5, 0) is  5\sqrt{2} ). So, the area  = side^2=(5\sqrt{2})^2=50 .

Hope it's clear.

Bunuel · Answer

BrushMyQuant · Answer

We can open |x| + |y| = 5 in four cases

Case 1 and 2 which is Quadrant I and Quadrant II
Quadrant I : x ≥ 0, y ≥ 0 => \|x\| = x, \| y\| = y => x + y = 5 => x = 0, y = 5 and y = 0, x = 5 So, the line will pass through the points (0,5) and (5,0) in Quadrant I	Quadrant II : x ≤ 0, y ≥ 0 => \|x\| = -x, \| y\| = y => -x + y = 5 => x = 0, y = 5 and y = 0, x = -5 So, the line will pass through the points (0,5) and (-5,0) in Quadrant II

Case 3 and 4 which is Quadrant III and Quadrant IV
Quadrant III : x ≤ 0, y ≤ 0 => \|x\| = -x, \| y\| = -y => -x - y = 5 => x = 0, y = -5 and y = 0, x = -5 So, the line will pass through the points (0,-5) and (-5,0) in Quadrant III	Quadrant IV : x ≥ 0, y ≤ 0 => \|x\| = x, \| y\| = -y => x - y = 5 => x = 0, y = -5 and y = 0, x = 5 So, the line will pass through the points (0,-5) and (5,0) in Quadrant IV

Attachment:

ABS-41-1.jpg [ 18.01 KiB | Viewed 2366 times ]

So, Area of the enclosed figure = Area of the Two triangles with base = (5 - (-5)) = 10 and Height = 5

Attachment:

ABS-41-2.jpg [ 18.64 KiB | Viewed 2357 times ]

=> Area of the enclosed figure = 2 * \(\frac{1}{2}\) * 10 * 5 = 50

So, Answer will be D
Hope it helps!

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If equation |x| + |y| = 5 encloses a certain region on the graph, what

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If equation |x| + |y| = 5 encloses a certain region on the graph, what

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