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Bunuel
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If equation |x| + |y| = 5 encloses a certain region on the graph, what 02 Apr 2019, 02:47
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D
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60% (01:19) correct 40% (01:24) wrong based on 895 sessions

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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

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Bunuel
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If equation |x| + |y| = 5 encloses a certain region on the graph, what 02 Apr 2019, 02:48
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Bunuel
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
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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:



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 \(\frac{\text{diagonal}^2}{2}=\frac{10^2}{2}=50\).


Answer: D


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georgethomps
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If equation |x| + |y| = 5 encloses a certain region on the graph, what 02 Apr 2019, 05:13
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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?
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If equation |x| + |y| = 5 encloses a certain region on the graph, what 02 Apr 2019, 05:55
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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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If equation |x| + |y| = 5 encloses a certain region on the graph, what 02 Apr 2019, 06:31
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m1033512
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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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.
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If equation |x| + |y| = 5 encloses a certain region on the graph, what 19 Jun 2019, 02:43
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georgethomps
m1033512
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

Posted from my mobile device
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.
Show more

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.
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If equation |x| + |y| = 5 encloses a certain region on the graph, what 26 Jan 2022, 09:28
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Why can't the area be computed as side^2 where each side equals 5 so area would be 25

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If equation |x| + |y| = 5 encloses a certain region on the graph, what 26 Jan 2022, 10:00
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Why can't the area be computed as side^2 where each side equals 5 so area would be 25

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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.
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If equation |x| + |y| = 5 encloses a certain region on the graph, what 26 Jan 2022, 11:29
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Why can't the area be computed as side^2 where each side equals 5 so area would be 25

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Similar questions to practice:



Hope it helps.
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If equation |x| + |y| = 5 encloses a certain region on the graph, what 30 May 2025, 20:31
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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
ABS-41-1.jpg [ 18.01 KiB | Viewed 1545 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
ABS-41-2.jpg [ 18.64 KiB | Viewed 1538 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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