If equation modX+modY=5 encloses a certain region on the coordinate plane, what is the area of that region?
5
10
25
50
100
Please explain this from the scratch as my coordinate geometry+graphs knowledge is absolutely nil.
much appreciated.
what is the area of that region?
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this one is pretty straight forward once you draw the figure on paper
X(max) = 5( i.e max value of x is 5)
Y(max) = 5
so, plot both X and Y on co-ordinate plane and join X and Y. You will get a straight line of length 5*sqrt(2)
Over here were took
x =+
y =+
now similarly, we have get 4 sets of straight lines( rather a square) of length 5*sqrt(2)
so, area = [5*sqrt(2)]^2
= 25*2 = 50
Hope this Helps
X(max) = 5( i.e max value of x is 5)
Y(max) = 5
so, plot both X and Y on co-ordinate plane and join X and Y. You will get a straight line of length 5*sqrt(2)
Over here were took
x =+
y =+
now similarly, we have get 4 sets of straight lines( rather a square) of length 5*sqrt(2)
so, area = [5*sqrt(2)]^2
= 25*2 = 50
Hope this Helps
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Excellent answer!
Let me see if I can reinforce this for myself and add to this:
1) If x(max)=5
and
y(max)=5
x= (5,0)
y= (0,5)
The distance between (0,0) and X(5,0) is 5.
The distance between (0,0) and Y(0,5) is 5.
Drawing a straight line between X and Y results in the distance being 5sqrt 2.
2) Next make X negative which results in (-5,0) but maintain Y as positive such that it is (0,5).
Once again the distance between X and Y is 5sqrt 2.
3) Next Make X and Y both negative such that X is (-5,0) and Y is (0,-5) resulting in a distance of 5sqrt2.
4) Lastly make X positive but Y negative such that X is (5,0) and Y is (0,-5). The distance is 5sqrt2.
Since all sides are equal, (5sqrt2), it forms a square. The area would be (5sqrt2)^2 = (5)^2 * (sqr2)^2 = 25 * 2 = 50.
Let me see if I can reinforce this for myself and add to this:
1) If x(max)=5
and
y(max)=5
x= (5,0)
y= (0,5)
The distance between (0,0) and X(5,0) is 5.
The distance between (0,0) and Y(0,5) is 5.
Drawing a straight line between X and Y results in the distance being 5sqrt 2.
2) Next make X negative which results in (-5,0) but maintain Y as positive such that it is (0,5).
Once again the distance between X and Y is 5sqrt 2.
3) Next Make X and Y both negative such that X is (-5,0) and Y is (0,-5) resulting in a distance of 5sqrt2.
4) Lastly make X positive but Y negative such that X is (5,0) and Y is (0,-5). The distance is 5sqrt2.
Since all sides are equal, (5sqrt2), it forms a square. The area would be (5sqrt2)^2 = (5)^2 * (sqr2)^2 = 25 * 2 = 50.
gmat740 wrote:this one is pretty straight forward once you draw the figure on paper
X(max) = 5( i.e max value of x is 5)
Y(max) = 5
so, plot both X and Y on co-ordinate plane and join X and Y. You will get a straight line of length 5*sqrt(2)
Over here were took
x =+
y =+
now similarly, we have get 4 sets of straight lines( rather a square) of length 5*sqrt(2)
so, area = [5*sqrt(2)]^2
= 25*2 = 50
Hope this Helps
- viju9162
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Hi Karan,
I am very poor at co-ordinate geometry. Can you guide me with any good resource/material to learn about co-ordinate geometry essential for GMAT?
Regards,
Viju
I am very poor at co-ordinate geometry. Can you guide me with any good resource/material to learn about co-ordinate geometry essential for GMAT?
Regards,
Viju
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- gmat740
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Hi, I would suggest you to refer to your High-School text books. they can really be helpful in developing the coordinate fundamentals and then you can move on to MGMAT math guides.
Hope this Helps
Hope this Helps
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modX + mody = 5 represents the following line equations:
x + y = 5 => (x/5) + (y/5) = 1;
x - y = 5 => (x/5) + (y/-5) = 1;
-x + y = 5 => (x/-5) + (y/5) = 1;
-x -y = 5 => (x/-5) + (y/-5) = 1;
A line equation of the type : (x/a) + (y/b) = 1 will intersect the X axis at (a,0) and the Y axis at (0,b) respectively.
So, the 4 equations will correspond to (5,0) (0,5) (-5,0) and (0, -5). This area represents a rhombus in the XY co-ordinate space
The area of a rhombus = product of the diagonals/2
both the diagonals of the region (a rhombus) are 10.
Therefore the area of the region = (10 X 10)/2 = 50.
x + y = 5 => (x/5) + (y/5) = 1;
x - y = 5 => (x/5) + (y/-5) = 1;
-x + y = 5 => (x/-5) + (y/5) = 1;
-x -y = 5 => (x/-5) + (y/-5) = 1;
A line equation of the type : (x/a) + (y/b) = 1 will intersect the X axis at (a,0) and the Y axis at (0,b) respectively.
So, the 4 equations will correspond to (5,0) (0,5) (-5,0) and (0, -5). This area represents a rhombus in the XY co-ordinate space
The area of a rhombus = product of the diagonals/2
both the diagonals of the region (a rhombus) are 10.
Therefore the area of the region = (10 X 10)/2 = 50.