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Coordinate geometry

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by raju232007 » Sun Sep 28, 2008 12:43 am
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2)sqrt(a^2)+sqrt(b^2) =sqrt(c^2) +sqrt(d^2)

I came across this ques in one of the mgmat cats.The answer is C.explanation sought.appreciate your help
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by 4meonly » Sun Sep 28, 2008 3:35 am
Could you post their solution here?
I get aswer B
because to make mentioned point equidistant from the origin they should be on the circumference with formula
x^2 + y^2 = R^2

May be question is not posted incorrectly, or I made a mistake
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by mals24 » Sun Sep 28, 2008 4:48 am
St 1 a/b=c/d
ad=bc
This st alone is insuff

St 2 sqrt(a)^2+sqrt(b)^2=sqrt(c)^2+sqrt(d)^2
i.e. a+b=c+d
This could have multiple solns: 4+1=4+1; 4+2=5+1
Hence INSUFF

Combining 1 & 2

Now we need to prove a^2+b^2=c^2+d^2

St 1 gives us ad=bc
St 2 gives us a+b=c+d---multiply this with b
(ab)+(b)^2=(cb)+(db)

ad=cb, substitute in the abv equ

(ab)+(b)^2=(ad)+(db)
b(a+b)=d(a+b)
b=d

since b=d
ad=bc
a=c

The x and y coords of the two points are same hence they are equidistant.

Answer is C.
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Re: Coordinate geometry

by ket_gmat » Sun Sep 28, 2008 5:49 am
raju232007 wrote:In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2)sqrt(a^2)+sqrt(b^2) =sqrt(c^2) +sqrt(d^2)

I came across this ques in one of the mgmat cats.The answer is C.explanation sought.appreciate your help


consider 1st statement

now to satisfy this condition selecting numbers
(a,b) = (1,2) & (c,d) = (3,6) so not equidistant from origin.
or if
(a,b) = (-1,-2) & (c,d) = (1,2) then equidistant from origin.
not sufficient.


consider 2nd statement.
selecting numbers
(a,b) = (1,2) & (c,d) = (1,2) then equidistant from origin.
but if
(a,b) = (1,2) & (c,d) = (2,1) so not equidistant from origin.
not sufficient.

Consider both together
to satisfy both conditions
I can select
(a,b) = (1,2) & (c,d) = (1,2)
or
(a,b) = (-1,-2) & (c,d) = (1,2)

rest of the number combinations are not satisfying both conditions.
looking at available options,We can say those two points are equidistant from origin.




:)
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by stop@800 » Sun Sep 28, 2008 7:59 am
C is the ans

we need

sqrt(a^2+b^2) = sqrt(c^2+d^2)

A:
ab = cd
we will not be able to get any value
what at the end we will get will be a simplified fraction

B
sqrt(a^2)+sqrt(b^2) = sqrt(c^2) +sqrt(d^2)
not suff
we can not get
sqrt(a^2+b^2) = sqrt(c^2+d^2)



A+b
sqrt(a^2)+sqrt(b^2) = sqrt(c^2) +sqrt(d^2)
a + b = c + d
square both sides
a2 + b2 + 2ab = c2 + d2 + 2cd

from A ab = cd so cancel both
a2 + b2 = c2 + d2
take square root on both sides
u have ur ans
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??

by TryHarder » Mon Sep 28, 2009 11:52 am
stop@800,
How did you get ab = cd, the provided condition says a/b = c/d!
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Re: Coordinate geometry

by maihuna » Mon Sep 28, 2009 1:11 pm
raju232007 wrote:In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2)sqrt(a^2)+sqrt(b^2) =sqrt(c^2) +sqrt(d^2)

I came across this ques in one of the mgmat cats.The answer is C.explanation sought.appreciate your help
Do you need a better explanation than MGMAT? Just follow their exp.
Charged up again to beat the beast :)
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by vaibhav.iit2002 » Tue Sep 29, 2009 10:04 am
mals24 wrote:St 1 a/b=c/d
ad=bc
This st alone is insuff

St 2 sqrt(a)^2+sqrt(b)^2=sqrt(c)^2+sqrt(d)^2
i.e. a+b=c+d
This could have multiple solns: 4+1=4+1; 4+2=5+1
Hence INSUFF

Combining 1 & 2

Now we need to prove a^2+b^2=c^2+d^2

St 1 gives us ad=bc
St 2 gives us a+b=c+d---multiply this with b
(ab)+(b)^2=(cb)+(db)

ad=cb, substitute in the abv equ

(ab)+(b)^2=(ad)+(db)
b(a+b)=d(a+b)
b=d

since b=d
ad=bc
a=c

The x and y coords of the two points are same hence they are equidistant.

Answer is C.

mals24,
b(a+b)=d(a+b)
b=d
we can say b=d only if a+b is not 0. how can we assure of that?
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by ardsouza » Wed Dec 21, 2011 9:30 am
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|

So we can cancel the third term out from LHS and RHS of equation 2. to get the desired equation
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by jainpiyushjain » Sun Mar 03, 2013 1:32 am
ardsouza wrote: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|

So we can cancel the third term out from LHS and RHS of equation 2. to get the desired equation
How did you arrive to the below equation ?
We need to prove a^2 + d^2 = c^2 + b^2
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by GMATGuruNY » Sun Mar 03, 2013 4:39 am
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?
(1) a/b = c/d
(2) √(a²) + √(b²) = √(c²) + √(d²)
Statement 1: a/b = c/d
If the equation is 1/2 = 1/2, then (1,2) and (1,2) are equidistant from the origin.
If the equation is 1/2 = 2/4, then (1,2) and (2,4) are not equidistant from the origin.
INSUFFICIENT.

Statement 2: √(a²) + √(b²) = √(c²) + √(d²)
√x² = |x|.
Rephrasing the statement, we get:
|a| + |b| = |c| + |d|.
If the equation is |0| + |2| = |0| + |2|, then (0,2) and (0,2) are equidistant from the origin.
If the equation is |0| + |2| = |1| + |1|, then (0,2) and (1,1) are not equidistant from the origin.
INSUFFICIENT.

Statements 1 and 2 combined:
Let a/b = c/d = k.
Then a=kb and c=kd.
Substituting a=kb and c=kd into |a|+|b| = |c|+|d|, we get:
|kb|+|b| = |kd|+|d|
|k+1| * |b| = |k+1| * |d|
|b| = |d|, implying that |a| = |c|.
Since the X VALUES in (a,b) and (c,d) are equidistant from the origin, and the Y VALUES in (a,b) and (c,d) are equidistant from the origin, the TWO POINTS THEMSELVES are equidistant from the origin.
SUFFICIENT.

The correct answer is C.

One take-away:
To evaluate each statement ON ITS OWN, I plugged in values.
To evaluate the two statements COMBINED, I applied algebra.
Many DS problems are best solved with this approach.
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