If the equation |x| + |y| = 6 encloses a certain region on the graph..

Question

If the equation  |x| + |y| = 6  encloses a certain region on the graph, what is the area of that region?

(A) 6 
(B) 12 
(C) 36 
(D) 72 
(E) 144

practicingadult · Answer

We get 4 lines from these equations, 
x + y = 6
x - y = 6
-x + y = 6
-x - y = 6

so we would get a square with sides 6\sqrt2, hence area 72

Krunaal · Answer

BrushMyQuant · Answer

We can open |x| + |y| = 6 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 = 6 => x = 0, y = 6 and y = 0, x = 6 So, the line will pass through the points (0,6) and (6,0) in Quadrant I	Quadrant II : x ≤ 0, y ≥ 0 => \|x\| = -x, \| y\| = y => -x + y = 6 => x = 0, y = 6 and y = 0, x = -6 So, the line will pass through the points (0,6) and (-6,0) 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 = 6 => x = 0, y = -6 and y = 0, x = -6 So, the line will pass through the points (0,-6) and (-6,0) in Quadrant III	Quadrant IV : x ≥ 0, y ≤ 0 => \|x\| = x, \| y\| = -y => x - y = 6 => x = 0, y = -6 and y = 0, x = 6 So, the line will pass through the points (0,-6) and (6,0) in Quadrant IV

Attachment:

ABS-40-1.jpg [ 17.2 KiB | Viewed 2177 times ]

So, Area of the enclosed figure = Area of the Two triangles with base = (6 - (-6)) = 12 and Height = 6

Attachment:

ABS-40-2.jpg [ 20.02 KiB | Viewed 2161 times ]

=> Area of the enclosed figure = 2 * \(\frac{1}{2}\) * 12 * 6 = 72

So, Answer will be D
Hope it helps!

Watch the following video to MASTER Absolute Value Problems

ektalakra · Answer

how will we determine that the coordinates will be 0-6 and not 2-4 or 4-2 or 1-5

Bunuel · Answer

It’s not that the “coordinates” are 0,6. What we’re really doing is finding the x- and y-intercepts of the equation |x| + |y| = 6.

To find the x-intercepts, set y = 0. Then |x| = 6, so x = 6 or x = -6. That means the graph cuts the x-axis at (6,0) and (-6,0).

To find the y-intercepts, set x = 0. Then |y| = 6, so y = 6 or y = -6. That means the graph cuts the y-axis at (0,6) and (0,-6).

These four points form a square tilted 45 degrees. The diagonal of the square is 12, and since the area of a square is (diagonal^2)/2, the area is (12^2)/2 = 72.

Please check the links to similar questions for more practice.

If the equation |x| + |y| = 6 encloses a certain region on the graph..

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If the equation |x| + |y| = 6 encloses a certain region on the graph..

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