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Equation |x| + |y| = 10 encloses a certain region on the coo

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Equation |x| + |y| = 10 encloses a certain region on the coo

by bhumika.k.shah » Sun Jan 24, 2010 6:24 am
Equation |x| + |y| = 10 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

OA D

Can someone please upload an attachment wherein these points could be plotted and how the figure would look like on the xy plane ?

Also various approaches neeeded to solve this sum would be much appreciated :D

thanks!
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by linkinpark » Sun Jan 24, 2010 6:57 am
Image

PS: Read 5 instead of 10 on the picture I've uploaded

from given equation for x= |5| and y = |5| value will be 10

so consider x and y = -5,5 both

the area covered will be a square with side = 10*root(2) so its area will be 200

**use formula to find distance between two points in our case it will be (5,5) to (-5,5) and (5,5) to (5,-5) which are basically top and right side of square

actually if you know co-ordinate system, its a logical thinking problem, solvable without thinking a lot
530->480->580
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by ajith » Sun Jan 24, 2010 7:09 am
---

Edited because of a mistakr
Last edited by ajith on Sun Jan 24, 2010 7:15 am, edited 1 time in total.
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by ajith » Sun Jan 24, 2010 7:13 am
ajith wrote:
bhumika.k.shah wrote:Equation |x| + |y| = 10 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

OA D

Can someone please upload an attachment wherein these points could be plotted and how the figure would look like on the xy plane ?

Also various approaches neeeded to solve this sum would be much appreciated :D

thanks!

The area enclosed by |x| + |y| = 10 is by
Lines x+y =10 x-y = 10, x +y =-10 and x-y = -10



Clearly x+y =10 and x+y = -10 are two parallel lines with distance of 20/sqrt(2) between them
x-y =10 and x-y = -10 are parallel and distance between them is 20/sqrt(2)

Explanation : Distance between two parallel lines ax+by =c and ax+by = d is c-d/sqrt(a^2+b^2)

these two sets are perpendicular making the enclosed space a square with side length =20/sqrt(2)

the area enclosed = 20*20 / (sqrt(2)*sqrt(2))= 200
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by ajith » Sun Jan 24, 2010 7:19 am
linkinpark wrote: PS: Read 5 instead of 10 on the picture I've uploaded

from given equation for x= |5| and y = |5| value will be 10

so consider x and y = -5,5 both

the area covered will be a square with side = 10*root(2) so its area will be 200

**use formula to find distance between two points in our case it will be (5,5) to (-5,5) and (5,5) to (5,-5) which are basically top and right side of square

actually if you know co-ordinate system, its a logical thinking problem, solvable without thinking a lot
The picture uploaded is not the right representation of lines

These lines are inclined at 45 degrees to the axes
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by bhumika.k.shah » Sun Jan 24, 2010 7:46 am
i took x+y = 10
and x + y =-10

and as per ur distance formula i got 10-(-10) = 20

now which two co-ordinates should i take in order to get sq rt 2 as the denominator ?

help me from thereon please
ajith wrote:
ajith wrote:
bhumika.k.shah wrote:Equation |x| + |y| = 10 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

OA D

Can someone please upload an attachment wherein these points could be plotted and how the figure would look like on the xy plane ?

Also various approaches neeeded to solve this sum would be much appreciated :D

thanks!

The area enclosed by |x| + |y| = 10 is by
Lines x+y =10 x-y = 10, x +y =-10 and x-y = -10



Clearly x+y =10 and x+y = -10 are two parallel lines with distance of 20/sqrt(2) between them
x-y =10 and x-y = -10 are parallel and distance between them is 20/sqrt(2)

Explanation : Distance between two parallel lines ax+by =c and ax+by = d is c-d/sqrt(a^2+b^2)

these two sets are perpendicular making the enclosed space a square with side length =20/sqrt(2)

the area enclosed = 20*20 / (sqrt(2)*sqrt(2))= 200
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by ajith » Sun Jan 24, 2010 7:55 am
bhumika.k.shah wrote:i took x+y = 10
and x + y =-10

and as per ur distance formula i got 10-(-10) = 20
The distance between two parallel lines ax+by =c and ax+by = d is

NOT c-d

instead it is

(c-d)/sqrt(a^2+b^2)

hence 20/Sqrt(2)
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by linkinpark » Sun Jan 24, 2010 7:58 am
ajith wrote:
linkinpark wrote:
The picture uploaded is not the right representation of lines

These lines are inclined at 45 degrees to the axes
would you please upload the corrected picture ...
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by ajith » Sun Jan 24, 2010 8:08 am
linkinpark wrote:
would you please upload the corrected picture ...
Attachments
Lines.doc
(27.5 KiB) Downloaded 165 times
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by bhumika.k.shah » Sun Jan 24, 2010 8:41 am
thankyou ajith!

nice explanation!
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by diebeatsthegmat » Wed Jun 02, 2010 6:20 pm
ajith wrote:
linkinpark wrote:
would you please upload the corrected picture ...
excuse me, i dont understand
if |x|+ |y|=10
so x=5, y=5,
x=-5 and y=5
x=5 y=-5
x=-5 y=-5
it could be. so we will have a square with 10 is the side
thus 10*10 =100
i dont understand why it could be 200. i dont understand why the side of the squre here turned into 200.?
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by nysnowboard » Wed Jun 02, 2010 9:23 pm
diebeatsthegmat wrote: excuse me, i dont understand
if |x|+ |y|=10
so x=5, y=5,
x=-5 and y=5
x=5 y=-5
x=-5 y=-5
it could be. so we will have a square with 10 is the side
thus 10*10 =100
i dont understand why it could be 200. i dont understand why the side of the squre here turned into 200.?
As an alternative to Ajith's excellent explantion:

The thing you are missing is that the sides of the square aren't parallel to the x and y axis. They are parallel to y =x and y = -x

Forget the values of 5, they are just arbitrary points on each side of the square. We want the vertexes. See my modification to Ajith's picture, attached.

abs(x) + abs(y) =10 will create four lines (which ultimately enclose a square, see Ajith's earlier attachment)

x+y =10 -------- y = -x +10
x+y =-10 ------- y = -x - 10
x-y = 10 -------- y = x -10
x-y = -10 ------- y = x +10

Instead of using his formula, you can picture the vertexes of the square at the x intercepts of 10,-10 and the y intercepts of 10, -10

Now each side is the hypotenuse of a 45-45-90 right triangle. Hence, the sides are in the ratio x : x : x sqrt(2)
Each side of the triangle is 10, so the hypotenuse is 10sqrt(2).

Remember, the hypotenuse of the triangle is a side of the square.

Area of a square is side^2 = (10sqrt(2))^2 =200.
Attachments
Lines-1.doc
(27.5 KiB) Downloaded 140 times
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