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absolute value + coordinate plane

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absolute value + coordinate plane

by tonebeeze » Tue Jan 04, 2011 12:30 pm
If equation: |x/2| (absolute value) + |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 don't understand the progression of this problem. I know |x| + |y| = 10 . But get stuck at that point. Please advise.

OA = [spoiler](D)[/spoiler]
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by clock60 » Tue Jan 04, 2011 1:36 pm
i`ll try to solve while my rationale is not elegant
solving this i operated with module, opening them and then constructing the figure
1) x/2>0, y/2>0 it means that x>0 and y>0 and returning to the initial equation
x/2+y/2=5. x+y=10, y=10-x, construct now this line, (x=10,y=0) may be first point and (x=0, y=10) second point
2) x/2<0, y/2<0 it means x<0, y<0 and
-(x/2)-(y/2)=5, x+y=-10, y=-10-x now construct this (x=0, y=-10) first point and (y=0 x=-10) second point
the same will be with cases
x>0, y<0 (x/2)-(y/2)=5 and at last
x<0, y>0 -(x/2)+(y/2)=5
out final figure will be square with coordinates (10,0) (0,10) (-10.0) (0,-10) or we have 4 equal right triangles with area
=1/2*10*10=50-of the one
and total 4*50=200
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by Tani » Tue Jan 04, 2011 2:09 pm
If x = 0, y = +10.
If y = 0, x = +10.

This forms a square, the diagonal of which is 20.
The diagonal of a square is equal to the side of the square times the square root of 2.
Therefore the side of your square is 20 divided by the square root of 2.

When you square the side you get 400/2 = 200
Tani Wolff
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