OG 12 - Problem solving -- Q 194 - The Beat The GMAT Forum - Expert GMAT Help & MBA Admissions Advice

OG 12 - Problem solving -- Q 194

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saifn: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Wed Nov 30, 2011 5:46 pm

by saifn » Mon Mar 26, 2012 5:00 am

This problem has been solved before on BTG but i would like a clarification of concept.

194.In the rectangular coordinate system above, the line y = x is the perpendicular bisector of segment AB (not shown), and the x-axis is the perpendicular bisector of segment BC (not shown). If the coordinates of point A are (2,3), what are the coordinates of point C ?
(A) (-3,-2)
(B) (-3,2)
(C) (2,-3)
(D) (3,-2)
(E) (2,3)

In the answer explanations it is given - Since the line y = x is the perpendicular bisector
of AB , B is the reflection of A through this line.

My question is are points reflections of each other only when their perpendicular bisectors are y=x or are they in all cases reflections of one another(i guess not),in which case if there was a y was not equal to x perpendicular bisector, how would you find the other point?

I hope i am making sense, i would be really grateful if someone could help me here.

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sam2304: Legendary Member; Posts: 1239; Joined: Tue Apr 26, 2011 6:25 amFollowed by: 26 members; Thanked: 233 times; GMAT Score:680

by sam2304 » Mon Mar 26, 2012 6:09 am

saifn wrote:My question is are points reflections of each other only when their perpendicular bisectors are y=x

Firstly your question is not very clear.

Try to rephrase it.

A and B are point reflections as line x = y divides the four quadrant into two equal halves with one mirroring the other. So they are reflections. The same line is the perpendicular bisector of the segment AB dividing the line into two equal parts. Since the two sides of line x = y are reflections B takes the value of (3,2), while A is (2,3).

are they in all cases reflections of one another(i guess not),in which case if there was a y was not equal to x perpendicular bisector, how would you find the other point?

The bold part does not make any sense. If y != x then we need more info to find the other point, as both the sides won't be reflections of one another. We can draw innumerable lines with y ! = x.

Please refer to the picture for better understanding.

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saifn: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Wed Nov 30, 2011 5:46 pm

by saifn » Tue Mar 27, 2012 12:02 am

I apologize for my badly framed question, but thanks a ton you have answered all my questions from the little you were able to comprehend.

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icanmakeit2bschool: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Tue Mar 06, 2012 8:33 pm; Thanked: 1 times

by icanmakeit2bschool » Tue Mar 27, 2012 10:48 pm

Saifn,

I think the coordinates for the point C is ( 3, -2 ).

saifn wrote:This problem has been solved before on BTG but i would like a clarification of concept.

194.In the rectangular coordinate system above, the line y = x is the perpendicular bisector of segment AB (not shown), and the x-axis is the perpendicular bisector of segment BC (not shown). If the coordinates of point A are (2,3), what are the coordinates of point C ?
(A) (-3,-2)
(B) (-3,2)
(C) (2,-3)
(D) (3,-2)
(E) (2,3)

In the answer explanations it is given - Since the line y = x is the perpendicular bisector
of AB , B is the reflection of A through this line.

My question is are points reflections of each other only when their perpendicular bisectors are y=x or are they in all cases reflections of one another(i guess not),in which case if there was a y was not equal to x perpendicular bisector, how would you find the other point?

I hope i am making sense, i would be really grateful if someone could help me here.

Like this post

ronnie1985: Legendary Member; Posts: 626; Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members; Location: Ahmedabad; Thanked: 31 times

by ronnie1985 » Wed Mar 28, 2012 8:33 am

A reflection of a point A(a,b) about a straight line y = mx+c can be obtained by following these steps:-

The point B(a1,b1) which is the reflection of point A(a,b) passes through a line perpendicular to the line y = mx+c. Let us assume that the equation of the perpendicular line is y = m1x+c1. Then m1*m = -1. Hence, m1 = -1/m. Since it passes through A(a,b) we can solve for m1 and c1. Then solve for the point of intersection of the lines y = mx+c and y = m1x+c1. Say the point is (x1,y1) then the answer can be obtained by:-

x1 = (a+a1)/2 and y1 = (b+b1)/2

Please revert back if more explanation is required.

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klmehta03: Senior | Next Rank: 100 Posts; Posts: 74; Joined: Sun Sep 05, 2010 3:11 pm; Thanked: 3 times

by klmehta03 » Thu Mar 29, 2012 3:11 am

IMO D. OA pls?

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[email protected]: Legendary Member; Posts: 934; Joined: Tue Nov 09, 2010 5:16 amFollowed by: 14 members; Location: AAMCHI MUMBAI LOCAL; Thanked: 63 times

by [email protected] » Fri Mar 30, 2012 12:19 am

The answer is between Options C and D, and I would go for the option D. well C is also not that wrong. What is the OA???

According to me the answer is D as see the words carefully of the question and see the diagram carefully.

line y = x is the perpendicular bisector of the Segment AB, that means the point where AB is getting intersected by line y = x should be equivalent to both the points A and B (just like a number line). IN that case Point B has the X-coordinate as 3 and not 2,

so even for the point C the X-Coordinate will be 3 and not 2.

Same thing applies to the X-axis bisecting the segment BC. hence the coordinates of the point C is (3,-2)...

Hope this helps... the confusion is only between points C and D as the point C should lie in the 4th quadrant and point B in the first quadrant...

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