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In the rectangular coordinate system, are the points (a, [#permalink]
 02 Jun 2009, 04:05
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In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin? (1) \frac{a}{b} =\frac{c}{d}(2) \sqrt{a^2} + \sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}
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Re: distance in rectangular coordinate system [#permalink]
02 Jun 2009, 07:44
mbaMission wrote: In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?
(1) a/b = c/d
(2) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2) (1) If a/b = c/d, a and b could be 100 and c and d could be 1 or a, b, c and d could be 1. NSF. (2) If (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2), then a+b = c+d. In this case, if a = 9 and b = 1, and c =d=5, a+b = c+d is true. but they have different distance.NSF. From 1 and 2, a must be equal to c and b must be equal to d. So Suff. Thats C.
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Re: distance in rectangular coordinate system [#permalink]
02 Jun 2009, 12:13
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin? (1) a/b = c/d This singnifies that the sign combination on both sides of = is the same. i.e. if one of the numbers on LHS is -ve , one of the numbers on RHS has to -ve as well. However since a,b / c,d can take any value , INSUFFICIENT. (2) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2) Since we do not know anything about signs of a,b / c,d - INSUFFICIENT. Using 1 and 2 together , using (1) the second statement drills down to a+b=c+d (the sign combination on both sides of = is the same.) Since a/b=c/d and a+b=c+d , (a, b) and (c, d) equidistant from the origin and infact represent the same point.
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Re: distance in rectangular coordinate system [#permalink]
02 Jun 2009, 18:23
amolsk11 wrote: In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?
(1) a/b = c/d
This singnifies that the sign combination on both sides of = is the same.
i.e. if one of the numbers on LHS is -ve , one of the numbers on RHS has to -ve as well.
However since a,b / c,d can take any value , INSUFFICIENT.
(2) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2)
Since we do not know anything about signs of a,b / c,d - INSUFFICIENT.
Using 1 and 2 together ,
using (1) the second statement drills down to
a+b=c+d (the sign combination on both sides of = is the same.)
Since a/b=c/d and a+b=c+d , (a, b) and (c, d) equidistant from the origin and infact represent the same point. Is there a way we can derive, using these two equations, that the two points are equidistant ?
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Re: distance in rectangular coordinate system [#permalink]
02 Jun 2009, 21:44
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goldeneagle94 wrote: amolsk11 wrote: In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?
(1) a/b = c/d
This singnifies that the sign combination on both sides of = is the same.
i.e. if one of the numbers on LHS is -ve , one of the numbers on RHS has to -ve as well.
However since a,b / c,d can take any value , INSUFFICIENT.
(2) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2)
Since we do not know anything about signs of a,b / c,d - INSUFFICIENT.
Using 1 and 2 together ,
using (1) the second statement drills down to
a+b=c+d (the sign combination on both sides of = is the same.)
Since a/b=c/d and a+b=c+d , (a, b) and (c, d) equidistant from the origin and infact represent the same point. Is there a way we can derive, using these two equations, that the two points are equidistant ? For the derivation part: a^2 + b^2 = c^2 + d^2 b^2[(a^2/b^2) + 1] = d^2[(c^2/d^2 +1)] --------> 1 from I a/b = c/d squaring both sides a^2/b^2 = c^2/d^2 putting above in 1 and simplifying b^2[(c^2/d^2) + 1] = d^2[(c^2/d^2 +1)] simplifying b^2 = d^2 or |b| = |d| => |a| = |c| therefore points are equidistant from (0,0) or Origin.
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Re: distance in rectangular coordinate system [#permalink]
03 Jun 2009, 07:24
in order for the points to be equidistant
a2 + b2 = c2+ d2 -----> eq X
1--> insufficent (has already been discussed)
2--> insufficent (has already been discussed)
combining both,
equation 2 can be written as a+b = c+d
eq1 a/b = c/d
a+b/b = c+d/d, from eq 1 from b=d ----3
similarly, b/a = d/c, a+b/a = d+c/c from eq 1 a = c ----4
plug in values of a, b in eq X... it satifies.
Hence points are equidistant. Answer is C
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Re: distance in rectangular coordinate system [#permalink]
08 Jun 2009, 23:29
GMAT TIGER wrote: mbaMission wrote: In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?
(1) a/b = c/d
(2) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2) (1) If a/b = c/d, a and b could be 100 and c and d could be 1 or a, b, c and d could be 1. NSF. (2) If (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2), then a+b = c+d. In this case, if a = 9 and b = 1, and c =d=5, a+b = c+d is true. but they have different distance.NSF. From 1 and 2, a must be equal to c and b must be equal to d. So Suff. Thats C. why from a^2 + b^2= c^2 + d^2 => a+b=c+d ?
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Re: distance in rectangular coordinate system [#permalink]
19 Aug 2009, 19:29
ngoctraiden1905 wrote: why from a^2 + b^2= c^2 + d^2 => a+b=c+d ? That is not so. What other posters said is that from (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2) => a+b=c+d That is actually incorrect. Should be |a|+|b|=|c|+|d|.
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Re: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 12:45
Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?
could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me
thanks for responses
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Re: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 13:06
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tinki wrote: Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?
could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me
thanks for responses a/b=c/d <---- Equation 1 |a|+|b|=|c|+|d| Add 1 to both sides of Equation 1 (a+b)/b=(c+d)/d Now a+b=c+d => 1/b=1/d Thus, b=d Similarly, b/a=d/c Adding 1 to both sides of above equation: (b+a)/a=(d+c)/c a+b=d+c => 1/a=1/c Therefore a=c
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Re: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 13:53
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tinki wrote: Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?
could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me
thanks for responses In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin? Distance between the point A (x,y) and the origin can be found by the formula: D=\sqrt{x^2+y^2}. So we are asked whether \sqrt{a^2+b^2}=\sqrt{c^2+d^2}? Or whether a^2+b^2=c^2+d^2? (1) \frac{a}{b}=\frac{c}{d} --> a=cx and b=dx, for some non-zero x. Not sufficient. (2) \sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2} --> |a|+|b|=|c|+|d|. Not sufficient. (1)+(2) From (1) a=cx and b=dx, substitute this in (2): |cx|+|dx|=|c|+|d| --> |x|(|c|+|d|)=|c|+|d| --> |x|=1 ( another solution |c|+|d|=0 is not possible as d in (1) given in denominator and can not be zero, so d\neq{0} --> |c|+|d|>0) --> now, as |x|=1 and a=cx and b=dx, then |a|=|c| and |b|=|d| --> square this equations: a^2=c^2 and b^2=d^2 --> add them: a^2+b^2=c^2+d^2. Sufficient. Answer: C.
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Re: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 14:27
Bunuel wrote: tinki wrote: Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?
could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me
thanks for responses In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin? Distance between the point A (x,y) and the origin can be found by the formula: D=\sqrt{x^2+y^2}. So we are asked whether \sqrt{a^2+b^2}=\sqrt{c^2+d^2}? Or whether a^2+b^2=c^2+d^2? (1) \frac{a}{b}=\frac{c}{d} --> a=cx and b=dx, for some non-zero x. Not sufficient. (2) \sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2} --> |a|+|b|=|c|+|d|. Not sufficient. (1)+(2) From (1) a=cx and b=dx, substitute this in (2): |cx|+|dx|=|c|+|d| --> |x|(|c|+|d|)=|c|+|d| --> |x|=1 ( another solution |c|+|d|=0 is not possible as d in (1) given in denominator and can not be zero, so d\neq{0} --> |c|+|d|>0) --> now, as |x|=1 and a=cx and b=dx, then |a|=|c| and |b|=|d| --> square this equations: a^2=c^2 and b^2=d^2 --> add them: a^2+b^2=c^2+d^2. Sufficient. Answer: C. your explanation is great as always ! thaaanks + Kudo
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Re: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 14:32
AmrithS wrote: tinki wrote: Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?
could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me
thanks for responses a/b=c/d <---- Equation 1 |a|+|b|=|c|+|d| Add 1 to both sides of Equation 1 (a+b)/b=(c+d)/d Now a+b=c+d => 1/b=1/d Thus, b=d Similarly, b/a=d/c Adding 1 to both sides of above equation: (b+a)/a=(d+c)/c a+b=d+c => 1/a=1/c Therefore a=c No that's not correct. |a|+|b|=|c|+|d| doesn't mean that a+b=c+d (consider a=b=1 and c=d=-1). So from |a|+|b|=|c|+|d| and a/b=c/d we can not derive that a=c and b=d. What we can derive is that |a|=|c| and |b|=|d|. Refer to my post above for complete solution.
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Re: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 20:36
Sorry! My Bad  Bunuel is right! Thanks for the explanation man....
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Re: distance in rectangular coordinate system [#permalink]
19 Feb 2011, 19:12
Entwistle wrote: tinki wrote: Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?
could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me
thanks for responses a/b=c/d <---- Equation 1 |a|+|b|=|c|+|d| Add 1 to both sides of Equation 1 (a+b)/b=(c+d)/d Now a+b=c+d => 1/b=1/d Thus, b=d Similarly, b/a=d/c Adding 1 to both sides of above equation: (b+a)/a=(d+c)/c a+b=d+c => 1/a=1/c Therefore a=c I liked your explanation, of course Bunuel was as helpful as ever too.
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Re: In the rectangular coordinate system, are the points (a, [#permalink]
21 Dec 2011, 10:34
Here's a simpler solution......
We need to prove a^2 + d^2 = c^2 + b^2
Simpler way is to rearrange statement 2 viz. |a| + |b| = |c| + |d| as
1. |a| - |d| = |c| - |b| and then square both sides. We get -
2. a^2 + d^2 - 2*|a|*|d| = c^2 + b^2 - 2*|c|*|b|.
Since ad = bc as per statement 1,
3. |ad| = |bc| => |a|.|d| = |b|.|c| (Rule -> Abs of Product = Product of Abs)
So we can cancel the third term out from both sides of equation 2. to get the desired equation
HTH
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Re: In the rectangular coordinate system, are the points (a, [#permalink]
27 Jul 2012, 21:49
Hi there, i was just wondering if the way i do it is correct!
Statement 1: insuff
Statement 2 : |a|+|b|=|c|+|d|
Our goal is to prove that a^2 + b^2 = c^2 + d^2
(1+2)
Square both sides in stmt 2.
We have a^2 + b^2 + 2|a||b| = c^2 + d^2 + 2|c||d| ----------- *
From one we know that a/b=c/d, therefore their LHS=RHS and therefore, this condition would allow us to cancel out 2|a||b| from LHS and 2|c||d| from equation *.
Please tell me that i am correct! =)
Reagan
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Re: In the rectangular coordinate system, are the points (a, [#permalink]
08 Aug 2012, 03:22
reagan wrote: Hi there, i was just wondering if the way i do it is correct!
Statement 1: insuff
Statement 2 : |a|+|b|=|c|+|d|
Our goal is to prove that a^2 + b^2 = c^2 + d^2
(1+2)
Square both sides in stmt 2.
We have a^2 + b^2 + 2|a||b| = c^2 + d^2 + 2|c||d| ----------- *
From one we know that a/b=c/d, therefore their LHS=RHS and therefore, this condition would allow us to cancel out 2|a||b| from LHS and 2|c||d| from equation *.
Please tell me that i am correct! =)
Reagan Absolutely, I found this approach much better. Infact I am wondering why we are targetting absolute values in the equation. (a+b)^2 = a^2+b^2+2ab [no absolute |a|, |b| needed] From 1) we know ab = cd 2) we know a + b = c + d. and hence (a+b)^2 = (c+d)^2 Combining 1) & 2) we can very well see that a^2+b^2 = c^2+d^2
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Re: In the rectangular coordinate system, are the points (a,
[#permalink]
08 Aug 2012, 03:22
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