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Right triangle ABO is drawn in the xy-plane, with OB as hypo

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Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

Got it off Manhattan GMAT blog(Challenge of the week)

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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 14 Jan 2014, 23:05
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mau5 wrote:
Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

Got it off Manhattan GMAT blog(Challenge of the week)


First off, draw the figure with whatever is given to you.
Attachment:
Ques3.jpg
Ques3.jpg [ 6.19 KiB | Viewed 8445 times ]

Vertex A has the right angle and could be at many places - the triangle could be 45-45-90 in which case it will be in the middle of O and B, it could be 30-60-90 triangle or many other variations. The area of all these triangles will be different.

(1) The x- and y-coordinates of all three points are non-negative integers.

The co-ordinates are non negative integers so A lies in the first quadrant. Also, if the co-ordinates of A are (a, b), the two sides will be given by:
\(OA^2 = a^2 + b^2\)
\(AB^2 = (15 - a)^2 + b^2\)
(a and b are positive integers)

Since OAB is a right triangle at A, \(OA^2 + AB^2 = OB^2\)
\(a^2 + b^2 + (15 - a)^2 + b^2 = 15^2\)
\(b^2 = (15 - a)*a\)
Now, we can quickly find out the values of a and b by putting in values of a from 1 to 7. Only a = 3 (or 12) gives us a perfect square on the right hand side.
So we will get 2 triangles a = 3, b = 6 or a = 12, b = 6. But note that both triangles will have the same area given by (1/2)*15*6 = 45 (1/2*OB*Altitude)
Hence statement 1 alone is sufficient.

(2) No two sides of the triangle have the same length.
This tells us that the triangle is not 45-45-90. Still the triangle can have many other angle measures such as 30-60-90 or 40-50-90 etc. The area will be different in different cases.

Answer (A)
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 25 Nov 2013, 21:59
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mau5 wrote:
Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

Got it off Manhattan GMAT blog(Challenge of the week)



With OB as hypotenuse for Right Angle Triangle, we can find the area of ABO by calculating the other 2 sides.

Length of Hypotenuse: 15 (Given)
The Other 2 sides of the triangle will be OA and AB.

from St1 we know that all three points are non-negative integers and so all points must be in quad 1.

We know for Pt O Co-ordinates are (0,0), Pt B (15,0).

for a right angle triangle, we have OA^2+AB^2= OB^2------> OA^2+AB^2 = 225 Because OB is 15.

OA and AB can be any points such 12,9 or 9,12 (Pythagorus triplets) and the Area of triangle will be 54.

So ans options can be A and D.

For St 2, Not sufficient...multiple possibilities.

Ans A.....
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 25 Nov 2013, 22:28
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WoundedTiger wrote:
mau5 wrote:
Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

Got it off Manhattan GMAT blog(Challenge of the week)



With OB as hypotenuse for Right Angle Triangle, we can find the area of ABO by calculating the other 2 sides.

Length of Hypotenuse: 15 (Given)
The Other 2 sides of the triangle will be OA and AB.

from St1 we know that all three points are non-negative integers and so all points must be in quad 1.

We know for Pt O Co-ordinates are (0,0), Pt B (15,0).

for a right angle triangle, we have OA^2+AB^2= OB^2------> OA^2+AB^2 = 225 Because OB is 15.

OA and AB can be any points such 12,9 or 9,12 (Pythagorus triplets) and the Area of triangle will be 54.

So ans options can be A and D.

For St 2, Not sufficient...multiple possibilities.

Ans A.....


Think again.

Hint: The co-ordinates have to be integral, not necessarily the length of the triangle. BTW, the area would not be 54.
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post Updated on: 26 Nov 2013, 01:03
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mau5 wrote:
WoundedTiger wrote:
mau5 wrote:
Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

Got it off Manhattan GMAT blog(Challenge of the week)



With OB as hypotenuse for Right Angle Triangle, we can find the area of ABO by calculating the other 2 sides.

Length of Hypotenuse: 15 (Given)
The Other 2 sides of the triangle will be OA and AB.

from St1 we know that all three points are non-negative integers and so all points must be in quad 1.

We know for Pt O Co-ordinates are (0,0), Pt B (15,0).

for a right angle triangle, we have OA^2+AB^2= OB^2------> OA^2+AB^2 = 225 Because OB is 15.

OA and AB can be any points such 12,9 or 9,12 (Pythagorus triplets) and the Area of triangle will be 54.

So ans options can be A and D.

For St 2, Not sufficient...multiple possibilities.

Ans A.....


Think again.

Hint: The co-ordinates have to be integral, not necessarily the length of the triangle. BTW, the area would not be 54.



You got me thinking.....Is this 600-700 level Q....I think it should be high..Anyways, I will give another try...

we know OA and AB intersect at right angle. Now let find the equation of these lines

For OA, let y= mx where m>0 as it is a postive slope

and Equation of AB be given by y-y1= a (x-x1) where y1, and x1 are 0 and 15 respectively or cor-ordinates of Point B.

Also note that because OA and AB intersect at right angle so a*m=-1 (If 2 lines are perpendicular then the product of their slopes is -1

For AB then we have y= a(x-15)

y= ax - 15a

Also y= mx so we have mx= ax-15a ----------> (m-a)x= -15 a

Now m= -1/a so we have (-1/a - a)x= -15 a ----------> (-1 -a^2)x= -15a^2

or x= 15 a^2/ (1 +a^2)

Now x has to be an integer value and only possible value for which x is an integer is a=-2 and we get x= 12

so y = mx or y=-1/ax------> y= 6

So Height of y= 6 and therefore Area of the triangle will be 1/2*15*6 = 45.....

Is this correct????

Note that the other possible value of a can be - \sqrt{2} and m =1/\sqrt{2}. Taking these values
We get y =10 and x=10/\sqrt{2} but x^2+y^2 is not equal to 225.

Also what is wrong with my way of doing this Q in first post. Why can't area by 54......

Lovely Question +1...........
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Originally posted by WoundedTiger on 25 Nov 2013, 22:58.
Last edited by WoundedTiger on 26 Nov 2013, 01:03, edited 3 times in total.
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 25 Nov 2013, 23:11
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WoundedTiger wrote:

or x= 15 a^2/ (1 +a^2)

Now x has to be an integer value and only possible value for which x is an integer is a=-2 and we get x= 12

so y = mx or y=-1/ax------> y= 6

So Height of y= 6 and therefore Area of the triangle will be 1/2*15*6 = 45.....



Absolutely. The area will be 45 sq. units. However, not to be a pain, but there is another value of a, for which we get integral value for x :

\(a=\frac{-1}{2}\), which gives x = 3. Note, that the value of y,however doesn't change in this case too: y=6 and as this is the height, the value of the area doesn't change.

It is more likely to be 700+, but I wanted to see the general trend before tagging with that difficulty level. Manhattan GMAT really create excellent questions. So kudos to them.
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 14 Jan 2014, 14:46
Great question thanks!

selected D for this one...

Can you develop on the statement 2 please?
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 14 Jan 2014, 20:10
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Paris75 wrote:
Great question thanks!

selected D for this one...

Can you develop on the statement 2 please?


St 2 basically says that the Right angle triangle is not isosceles because in Isosceles (45:45:90Right triangle the sides will be a:a:a\(\sqrt{2}\)

The triangle can be a 30:60:90 or x:y:90 where x and y can take any values but the sum of x+y=90
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 15 Jan 2014, 00:02
WoundedTiger wrote:
Paris75 wrote:
Great question thanks!

selected D for this one...

Can you develop on the statement 2 please?


St 2 basically says that the Right angle triangle is not isosceles because in Isosceles (45:45:90Right triangle the sides will be a:a:a\(\sqrt{2}\)

The triangle can be a 30:60:90 or x:y:90 where x and y can take any values but the sum of x+y=90


Misunderstood statement 2! Thanks! Good question!
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 16 Oct 2015, 01:24
VeritasPrepKarishma wrote:
mau5 wrote:
Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

Got it off Manhattan GMAT blog(Challenge of the week)


First off, draw the figure with whatever is given to you.
Attachment:
Ques3.jpg

Vertex A has the right angle and could be at many places - the triangle could be 45-45-90 in which case it will be in the middle of O and B, it could be 30-60-90 triangle or many other variations. The area of all these triangles will be different.

(1) The x- and y-coordinates of all three points are non-negative integers.

The co-ordinates are non negative integers so A lies in the first quadrant. Also, if the co-ordinates of A are (a, b), the two sides will be given by:
\(OA^2 = a^2 + b^2\)
\(AB^2 = (15 - a)^2 + b^2\)
(a and b are positive integers)

Since OAB is a right triangle at A, \(OA^2 + AB^2 = OB^2\)
\(a^2 + b^2 + (15 - a)^2 + b^2 = 15^2\)
\(b^2 = (15 - a)*a\)
Now, we can quickly find out the values of a and b by putting in values of a from 1 to 7. Only a = 3 (or 12) gives us a perfect square on the right hand side.
So we will get 2 triangles a = 3, b = 6 or a = 12, b = 6. But note that both triangles will have the same area given by (1/2)*15*6 = 45 (1/2*OB*Altitude)
Hence statement 1 alone is sufficient.

(2) No two sides of the triangle have the same length.
This tells us that the triangle is not 45-45-90. Still the triangle can have many other angle measures such as 30-60-90 or 40-50-90 etc. The area will be different in different cases.

Answer (A)


Hello VeritasPrepKarishma
could you please explain how do you get \(b^2 = (15 - a)*a\) equation?
tnx
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 17 Oct 2015, 22:37
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

If we modify the original condition, the base is 15, so we only need to know the height. In other words, there is only have one variable and we only require one equation in order to solve the question. The conditions provide us with 2 equations, so there is high chance (D) will be our asnwer.
Looking at condition 1, there is only A(9,12) or A(12,9). (The values which are non-negative integers are 3:4:5=9:12:15) so this is a sufficient condition,
For condition 2, there are too many situations possible, so this is insufficient, and the answer therefore becomes (A).
This type of question is not tested in nowadays GMAT.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 18 Oct 2015, 23:25
1
amirzohrevand wrote:
VeritasPrepKarishma wrote:
mau5 wrote:
Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

Got it off Manhattan GMAT blog(Challenge of the week)


First off, draw the figure with whatever is given to you.
Attachment:
Ques3.jpg

Vertex A has the right angle and could be at many places - the triangle could be 45-45-90 in which case it will be in the middle of O and B, it could be 30-60-90 triangle or many other variations. The area of all these triangles will be different.

(1) The x- and y-coordinates of all three points are non-negative integers.

The co-ordinates are non negative integers so A lies in the first quadrant. Also, if the co-ordinates of A are (a, b), the two sides will be given by:
\(OA^2 = a^2 + b^2\)
\(AB^2 = (15 - a)^2 + b^2\)
(a and b are positive integers)

Since OAB is a right triangle at A, \(OA^2 + AB^2 = OB^2\)
\(a^2 + b^2 + (15 - a)^2 + b^2 = 15^2\)
\(b^2 = (15 - a)*a\)
Now, we can quickly find out the values of a and b by putting in values of a from 1 to 7. Only a = 3 (or 12) gives us a perfect square on the right hand side.
So we will get 2 triangles a = 3, b = 6 or a = 12, b = 6. But note that both triangles will have the same area given by (1/2)*15*6 = 45 (1/2*OB*Altitude)
Hence statement 1 alone is sufficient.

(2) No two sides of the triangle have the same length.
This tells us that the triangle is not 45-45-90. Still the triangle can have many other angle measures such as 30-60-90 or 40-50-90 etc. The area will be different in different cases.

Answer (A)


Hello VeritasPrepKarishma
could you please explain how do you get \(b^2 = (15 - a)*a\) equation?
tnx



\(a^2 + b^2 + (15 - a)^2 + b^2 = 15^2\)

\(2b^2 = 15^2 - a^2 - (15 - a)^2\)

\(2b^2 = (15 - a)(15 + a) - (15 - a)^2\)

\(2b^2 = (15 - a)[(15 + a ) - (15 - a)]\)

\(2b^2 = (15 - a)[15 + a - 15 + a]\)

\(2b^2 = (15 - a) * 2a\)

\(b^2 = (15 - a)*a\)
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 08 Nov 2015, 16:00
how many out of you guys get the answer within 1 minutes 30 seconds?

I can get quite fast:

y/x=tga=15-x/y===>trigonometry helps a lot in these situations

then

y^2=x(15-x)

but it does take me like five minutes to make sure there's only one integer pair
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 11 May 2016, 10:10
MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

If we modify the original condition, the base is 15, so we only need to know the height. In other words, there is only have one variable and we only require one equation in order to solve the question. The conditions provide us with 2 equations, so there is high chance (D) will be our asnwer.
Looking at condition 1, there is only A(9,12) or A(12,9). (The values which are non-negative integers are 3:4:5=9:12:15) so this is a sufficient condition,
For condition 2, there are too many situations possible, so this is insufficient, and the answer therefore becomes (A).
This type of question is not tested in nowadays GMAT.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.





Can someone verify that this type of question isnt tested nowadays in gmat?
This question was so involving... I don't think there is a way for one to possibly work through all this Math in under 5 minutes. I would be absolutely baffled if a question like this is testable. Any thoughts?

Also, I don't believe this explanation is correct because the question states the coordinates of point A are integer values, not the lengths of the legs themselves.
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 11 May 2016, 19:58
mattyahn wrote:
MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

If we modify the original condition, the base is 15, so we only need to know the height. In other words, there is only have one variable and we only require one equation in order to solve the question. The conditions provide us with 2 equations, so there is high chance (D) will be our asnwer.
Looking at condition 1, there is only A(9,12) or A(12,9). (The values which are non-negative integers are 3:4:5=9:12:15) so this is a sufficient condition,
For condition 2, there are too many situations possible, so this is insufficient, and the answer therefore becomes (A).
This type of question is not tested in nowadays GMAT.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.





Can someone verify that this type of question isnt tested nowadays in gmat?
This question was so involving... I don't think there is a way for one to possibly work through all this Math in under 5 minutes. I would be absolutely baffled if a question like this is testable. Any thoughts?

Also, I don't believe this explanation is correct because the question states the coordinates of point A are integer values, not the lengths of the legs themselves.


The question is not out of scope, but certainly very unlikely and in the 750 - 800 range. If you are killing everything in your way, you might get to see something like this. But I must point out that it involves some painful recursive action - check what value(s) of a gives you a perfect square. That isn't GMAT's style.
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 30 Sep 2016, 02:27
Could someone explain to me why this cannot be calculated simply using the pythagoras theorem. (A) is necessary because otherwise the coordinates could be fractions which will cause the area to be unsolvable, otherwise I can't see why it's important that it lies in quadrant 1. It's only important that the coordinates are integers? Anyway I hope this make sense
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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New post 26 Nov 2016, 11:21
JimminyCricket wrote:
Could someone explain to me why this cannot be calculated simply using the pythagoras theorem. (A) is necessary because otherwise the coordinates could be fractions which will cause the area to be unsolvable, otherwise I can't see why it's important that it lies in quadrant 1. It's only important that the coordinates are integers? Anyway I hope this make sense


If you apply the pythogaros theorem and with (x,y) as the point A, the locus of all points that can satisfy the eqn will be a circle with centre at (7.5,0) with radius 7.5. So for any integral value of x that can satisfy the equation we will have 2 y values, one positive and the other negative. Thus, the mathematical solution if x and y can be anything will have 2 solutions, one in the 1st Quadrant and the other in the 4th Quadrant.

I know it's hard to imagine, but maths helps us see what we can't visualize. Hope I helped. All the best.

+1 kudos if you found it helpful.
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Re: Right triangle ABO is drawn in the xy-plane, with OB as hypo  [#permalink]

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