## GMAT Prep - Geometry (2)

Math questions from mba.com and GMAT Prep software
Khush
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### Re: GMAT Prep - Geometry (2)

sure!
RonPurewal
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### Re: GMAT Prep - Geometry (2)

.
SamikB952
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### Re: GMAT Prep - Geometry (2)

Hey guys,
On a problem like this, when you see O, and x- and y- axis, is it safe to assume that O is the origin (0,0). It isn't explicitly stated, so I wasn't sure if I could assume that.
RonPurewal
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### Re: GMAT Prep - Geometry (2)

gmac will not use "O" for any other point.

because "O" is traditionally used for the origin, that would constitute trickery; there are no trick questions on this exam.
RonPurewal
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### Re: GMAT Prep - Geometry (2)

more importantly—

if you thought "O" might not be the origin... what other point did you think it might indicate?
there are no other points shown in that neighborhood of the diagram. so, even if you're not aware of the convention i mentioned above, there's nothing else that "O" could possibly indicate here.
ajaym8
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### Re: GMAT Prep - Geometry (2)

Hi instructors,

1. Apart from Ron's approach (it was fantastic), is there any other way this problem can be done ?
2. Since we know that it's a 45-45-90 isosceles right triangle, doesn't that imply that PQ is parallel to X-axis ? Let me elaborate on why I am thinking this.
I imagined PQ as not parallel to x-axis. Somewhat slanting line PQ is. But that would change the 45 degree angle. No ?
3. This question occurred to me when I was trying to find out alternative approaches.
Consider a straight line AB with A(x1,y1) and B(x2,y2). If I know the length of AB =m, and I know the values of x1,y1, is there a way I can find out values of
variables x2,y2 ? I google-searched this. I found the distance formula everywhere but not the answer to this question that cropped up in my mind.

Thanks,
ajaym8
ajaym8
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### Re: GMAT Prep - Geometry (2)

Ok. I got the answer to my first question. Another way using the property of slopes of lines. Still, any other ways ?

ajaym8 wrote:Hi instructors,

1. Apart from Ron's approach (it was fantastic), is there any other way this problem can be done ?
2. Since we know that it's a 45-45-90 isosceles right triangle, doesn't that imply that PQ is parallel to X-axis ? Let me elaborate on why I am thinking this.
I imagined PQ as not parallel to x-axis. Somewhat slanting line PQ is. But that would change the 45 degree angle. No ?
3. This question occurred to me when I was trying to find out alternative approaches.
Consider a straight line AB with A(x1,y1) and B(x2,y2). If I know the length of AB =m, and I know the values of x1,y1, is there a way I can find out values of
variables x2,y2 ? I google-searched this. I found the distance formula everywhere but not the answer to this question that cropped up in my mind.

Thanks,
ajaym8
RonPurewal
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### Re: GMAT Prep - Geometry (2)

2. Since we know that it's a 45-45-90 isoscel
es right triangle, doesn't that imply that PQ is parallel to X-axis ? Let me elaborate on why I am thinking this.

not necessarily.
such a triangle CAN have side PQ parallel to the x-axis -- but, you can also imagine "pinning down" the vertex at point O, and then spinning the rest of the triangle around that point. if you do this, every other possible location of the triangle will have that side at an oblique angle to the x-axis.

I imagined PQ as not parallel to x-axis. Somewhat slanting line PQ is.

this ^^ is correct; PQ is not horizontal.
this does not prohibit the triangle from being 45°-45°-90°, as explained above.
RonPurewal
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### Re: GMAT Prep - Geometry (2)

3. This question occurred to me when I was trying to find out alternative approaches.
Consider a straight line AB with A(x1,y1) and B(x2,y2). If I know the length of AB =m, and I know the values of x1,y1, is there a way I can find out values of variables x2,y2 ?

...no, that's not possible.

you don't need any actual knowledge to ascertain this, either -- you can just think about the situation with everyday common sense.
your given information consists of only (i) a fixed point, and (ii) a fixed distance from that point.
clearly, there is not just ONE point at a given distance from the original point! (...in fact, there is an entire circle made up of such points) therefore there will most certainly not be a uniquely defined answer to this.
MarenM892
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### Re: GMAT Prep - Geometry (2)

Hey Ron! Could you please check the below solution and explain, why we are able to set M + √3 and M + 1 equal to each other?

A friend told me the below solution and it produces the correct answer - but I am unsure how he gets to the first step and if the assumptions he is making are valid.

Thanks a lot!

1. Set equal distances

O - (-√3) = O +1
2*O = 1- √3
O = ( 1- √3 ) / 2

2. Plug into equation for s

s = O + (O - (-√3)
s = ( 1 - √3 ) / 2 + ( 1 - √3 ) / 2 + √3
s = 1 - √3 + √3
s = 1
Sage Pearce-Higgins
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### Re: GMAT Prep - Geometry (2)

I'm having trouble understanding your question here.

why we are able to set M + √3 and M + 1 equal to each other?

Which M are you referring to?

O - (-√3) = O +1
2*O = 1- √3
O = ( 1- √3 ) / 2

What does "O" stand for in this equation?