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Source: Beat The GMAT — Data Sufficiency |
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samirpandeyit62
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The ans should be D
stmt 1 : y= x this line will make 45 deg angle with x & y axis &will have pts like (1,1) (2,2) etc
now vertical distance of pt (1,2) from (1,1) is 1 unit, now if join these pts then the angle made with y=x would be 45 degs (alternate angles) so
Now actual distance of pt is perpendicular dropped from pt (1,2) to y=x ,so one angle is 45 degs, one 90 degs so third is 45 degs,
similarly vertical distance of pt (2,1) from (2,2) is 1 unit, now if join these pts then the angle made with y=x would be 45 degs (alternate angles +vrtically opposite) so Now actual distance of pt is perpendicular dropped from pt (2,1) to y=x ,so one angle is 45 degs, one 90 degs so third is 45 degs,
now these two traingles would be congruent as all angles are same and 1 side is same i.e 1unit so lenght of perpendiculars would be same
henec pts are equidistant
so SUFF (cannot be clsoer)
stnt 2: y=-x will also make angles of 45 deg with x & y axis, and will have pts like (1,-1) (2,-2) etc
using above logic we can again prove that the pts are equidistant
henec SUFF
Pls confirm the answer
stmt 1 : y= x this line will make 45 deg angle with x & y axis &will have pts like (1,1) (2,2) etc
now vertical distance of pt (1,2) from (1,1) is 1 unit, now if join these pts then the angle made with y=x would be 45 degs (alternate angles) so
Now actual distance of pt is perpendicular dropped from pt (1,2) to y=x ,so one angle is 45 degs, one 90 degs so third is 45 degs,
similarly vertical distance of pt (2,1) from (2,2) is 1 unit, now if join these pts then the angle made with y=x would be 45 degs (alternate angles +vrtically opposite) so Now actual distance of pt is perpendicular dropped from pt (2,1) to y=x ,so one angle is 45 degs, one 90 degs so third is 45 degs,
now these two traingles would be congruent as all angles are same and 1 side is same i.e 1unit so lenght of perpendiculars would be same
henec pts are equidistant
so SUFF (cannot be clsoer)
stnt 2: y=-x will also make angles of 45 deg with x & y axis, and will have pts like (1,-1) (2,-2) etc
using above logic we can again prove that the pts are equidistant
henec SUFF
Pls confirm the answer
Regards
Samir
Samir
Samir, for stmt 2 lets assume points (3,-3) and (-3,3)
Distance between (3,-3) and (1,2) = sqrt(29)
Distance between (3,-3) and (2,1) = sqrt(17)
Distance between (-3,3) and (1,2) = sqrt(17)
Distance between (-3,3) and (2,1) = sqrt(29)
so, ans should be A
Distance between (3,-3) and (1,2) = sqrt(29)
Distance between (3,-3) and (2,1) = sqrt(17)
Distance between (-3,3) and (1,2) = sqrt(17)
Distance between (-3,3) and (2,1) = sqrt(29)
so, ans should be A
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samirpandeyit62
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oh sorry I interpreted the Q wrongly, I thought it is required to find the distnce of pts (1,2) & (2,1) from the lines (was watching 20-20 Cricket Worldcup & typing ocassionaly so was not very attentive)
Then IMO the ans should be C
stmt 1: as if pt A is to the left of pt (1,2) then it is closer to it
if it is to the right of (2,1) then its is closer to it, all depends upon the location of pt
NOT SUFF
stmt 2: as if pt A is to the left of pt (1,2) say in second quadrant then it is closer to it
if it is below & right of (2,1) say in the 4th quadrant then its is closer to it, all depends upon the location of pt
NOT SUFF
togther we can say pt A is origin (0,0) which would be equidistant from (1,2) & (2,1) SUFF
ans should be C
Then IMO the ans should be C
stmt 1: as if pt A is to the left of pt (1,2) then it is closer to it
if it is to the right of (2,1) then its is closer to it, all depends upon the location of pt
NOT SUFF
stmt 2: as if pt A is to the left of pt (1,2) say in second quadrant then it is closer to it
if it is below & right of (2,1) say in the 4th quadrant then its is closer to it, all depends upon the location of pt
NOT SUFF
togther we can say pt A is origin (0,0) which would be equidistant from (1,2) & (2,1) SUFF
ans should be C
Last edited by samirpandeyit62 on Tue Sep 11, 2007 10:27 pm, edited 1 time in total.
Regards
Samir
Samir
for stmt 1 lets assume points (3,3) and (-3,-3)
Distance between (3,3) and (1,2) = sqrt(5)
Distance between (3,3) and (2,1) = sqrt(5)
Distance between (-3,-3) and (1,2) = sqrt(41)
Distance between (-3,-3) and (2,1) = sqrt(41)
In any case the points are equidistant
Distance between (3,3) and (1,2) = sqrt(5)
Distance between (3,3) and (2,1) = sqrt(5)
Distance between (-3,-3) and (1,2) = sqrt(41)
Distance between (-3,-3) and (2,1) = sqrt(41)
In any case the points are equidistant
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samirpandeyit62
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Hi kajcha,
I'm not able to concentrate much but I evaluated it yes A should be the answer, a simple explanation join the pts with A & since perpendicular distance is same & the perpedicular makes an 90 so if we make a traingle joining the pts with A and dropping perpendiculars to the the line from the pts , we would have two congruent traingles (one side is the line y=x) hence pts will be equidistant. so ans A
I'm not able to concentrate much but I evaluated it yes A should be the answer, a simple explanation join the pts with A & since perpendicular distance is same & the perpedicular makes an 90 so if we make a traingle joining the pts with A and dropping perpendiculars to the the line from the pts , we would have two congruent traingles (one side is the line y=x) hence pts will be equidistant. so ans A
Regards
Samir
Samir
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samirpandeyit62
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Hi All,
Sorry for the confusing posts yesterday (except the last one). I would like to present a way to solve this quickly,
If you analyse that line y=x will equally divide the first & 3rd quadrants, as it would make angles of 45 deg with both the x & y axis's.
Now if u make any figure say a trapazium by joining the the pts (1,2) & (2,1) with the y axis & x repectively & also with the line y=x, then we can say that these two figures are mirror images of each other across the line y=x (use symmetry)
hence we can say pts (1,2) & (2,1) are symmetrical across y=x, so they are equidistant from y=x perpendicularly, (so if A is perpendicular to these pts then distance is equal)
Now A can anywhere alse on y=x, in that case pt symmetrical to A across y=x is A itself so now if we join (1,2) & (2,1) with A & y=x (perpendicularly say this is pt B) to create one traingle each with both the pts joined to A & B we can say that all three pts are symmetrical across y=x(i.e. A sym to A on same line y=x, B is sym to B & as we showed above (1,2) & (2,1) ) so the triangles will be mirror images of each other hence both pts will be equidistant from each other.
The explanation may seem lenghty but on the scratchpaper + ur mind it will only take few secs & no calcuation to say (1,2) & (2,1) are equidistant from A.
Sorry for the confusing posts yesterday (except the last one). I would like to present a way to solve this quickly,
If you analyse that line y=x will equally divide the first & 3rd quadrants, as it would make angles of 45 deg with both the x & y axis's.
Now if u make any figure say a trapazium by joining the the pts (1,2) & (2,1) with the y axis & x repectively & also with the line y=x, then we can say that these two figures are mirror images of each other across the line y=x (use symmetry)
hence we can say pts (1,2) & (2,1) are symmetrical across y=x, so they are equidistant from y=x perpendicularly, (so if A is perpendicular to these pts then distance is equal)
Now A can anywhere alse on y=x, in that case pt symmetrical to A across y=x is A itself so now if we join (1,2) & (2,1) with A & y=x (perpendicularly say this is pt B) to create one traingle each with both the pts joined to A & B we can say that all three pts are symmetrical across y=x(i.e. A sym to A on same line y=x, B is sym to B & as we showed above (1,2) & (2,1) ) so the triangles will be mirror images of each other hence both pts will be equidistant from each other.
The explanation may seem lenghty but on the scratchpaper + ur mind it will only take few secs & no calcuation to say (1,2) & (2,1) are equidistant from A.
Regards
Samir
Samir
