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Grockit tough problem

by arora007 » Fri Jan 28, 2011 6:26 am
In the rectangular coordinate system to the left, the line y = -x is the perpendicular bisector of segment RS (not shown) and the y-axis is the perpendicular bisector of segment ST (not shown). If the coordinates of point R are (-3, 1), what are the coordinates of point T ?

A (-1, 3)
B (-1, -3)
C (3, -1)
D (1, 3)
E (3, 1)

OA D

please help me with an easy explanation.
Also request you to help me with the simple approach to such problems.
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by GMATGuruNY » Fri Jan 28, 2011 6:42 am
arora007 wrote:In the rectangular coordinate system to the left, the line y = -x is the perpendicular bisector of segment RS (not shown) and the y-axis is the perpendicular bisector of segment ST (not shown). If the coordinates of point R are (-3, 1), what are the coordinates of point T ?

A (-1, 3)
B (-1, -3)
C (3, -1)
D (1, 3)
E (3, 1)

OA D

please help me with an easy explanation.
Also request you to help me with the simple approach to such problems.
Just draw a picture of what is being described. A perpendicular bisector:

-- intersects at the midpoint
-- forms a right angle

Here's my rudimentary drawing:

Image

Looking at the drawing above, we can see that T must have a positive x coordinate and a positive y coordinate. Since point T is higher than point R, the y coordinate of point T must be greater than 1. Only answer choice D works.

The correct answer is D.
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by stormier » Fri Jan 28, 2011 7:47 am
Mitch's solution is the best way to solve such problem. Visualization is the key. Although, you could solve the problem without a drawing.

line y=-x bisects RS ==> RS is perpendicular to line y=-x. Thus slope of RS = +1

equation of RS becomes y=x +c, and since the line passes through R (-3,1), the value of c = 4

equation of line RS is y = x +4

Solve for intersection point of RS and line y=-x to obtain (-2,2) as the point of intersection. This point of intersection is the midpoint of segment RS

R (-3,1)

midpoint (-2,2); midpoint of (x1,y1) and (x2,y2) is [(x1+x2)/2, (y1+y2)/2]

Thus S is (-1,3)

Now, segment ST is bisected by the y-axis ==> ST is parallel to x axis

So it intersects with y axis at (0,3)

point (0,3) is the midpoint of S(-1,3) and T(x,y)

thus T is (1,3)

Arora007 - thanks for pointing out the mistake. fixed.
Last edited by stormier on Sat Jan 29, 2011 10:02 am, edited 1 time in total.
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by arora007 » Fri Jan 28, 2011 8:54 am
Thanks Mitch... you are a GEM!! You are a great great help to all of us students...

stormier: "line y=-x to obtain (-2,1) as the point of intersection" this point should be (-2,2)
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by thebigkats » Thu Mar 03, 2011 5:01 pm
We have a line represented by y = -x (SLOPE m= -1)

RS is perpendicular to it, so SLOPE = -1/m = -1/-1 = 1

Lines with positive slope means that both x and y values grow as we go from beginning of segment (R) to the end of segment (S).

Coord of points R = (-3, 1). So if S is represented by (x,y) then x > -3 and y > 1.

Moreover ST is perpendicularly bisected by y-axis, meaning that it is parallal to x axis as well as T (x) = +ve

This means that if T = (x2, y2) then x2 > -3 and +ve and y2 > y >1

From answers choices, the only x value that fits the mold are C, D and E
the only y value that fits the mold is A and D

Hence D
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by leumas » Mon Dec 19, 2011 11:34 pm
GMATGuruNY wrote:
arora007 wrote:In the rectangular coordinate system to the left, the line y = -x is the perpendicular bisector of segment RS (not shown) and the y-axis is the perpendicular bisector of segment ST (not shown). If the coordinates of point R are (-3, 1), what are the coordinates of point T ?

A (-1, 3)
B (-1, -3)
C (3, -1)
D (1, 3)
E (3, 1)

OA D

please help me with an easy explanation.
Also request you to help me with the simple approach to such problems.
Just draw a picture of what is being described. A perpendicular bisector:

-- intersects at the midpoint
-- forms a right angle

Here's my rudimentary drawing:

Image

Looking at the drawing above, we can see that T must have a positive x coordinate and a positive y coordinate. Since point T is higher than point R, the y coordinate of point T must be greater than 1. Only answer choice D works.

The correct answer is D.
Hi Mitch, can you please re post the image again? It is not working!
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by ankush123251 » Mon Dec 19, 2011 11:47 pm
My approach,

R = (-3,1)

Since y = -x is perpendicular bisector of RS point S will be symmetrical about line y = -x.Thus point S will be (-1,3).
Again Y axis is perpendicular bisector of ST.Thus,T will be symmetrical to S about Y axis.
Thus,
T = (1,3)
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by ronnie1985 » Thu Apr 26, 2012 10:46 am
Draw a figure for yourself
One is attached for your understanding
Attachments
snapshot.jpg
Co-ordinate Geometry
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by karthikpandian19 » Sun Jun 03, 2012 11:14 pm
Mitch, the drawing was so self explanatory
GMATGuruNY wrote:
arora007 wrote:In the rectangular coordinate system to the left, the line y = -x is the perpendicular bisector of segment RS (not shown) and the y-axis is the perpendicular bisector of segment ST (not shown). If the coordinates of point R are (-3, 1), what are the coordinates of point T ?

A (-1, 3)
B (-1, -3)
C (3, -1)
D (1, 3)
E (3, 1)

OA D

please help me with an easy explanation.
Also request you to help me with the simple approach to such problems.
Just draw a picture of what is being described. A perpendicular bisector:

-- intersects at the midpoint
-- forms a right angle

Here's my rudimentary drawing:

Image

Looking at the drawing above, we can see that T must have a positive x coordinate and a positive y coordinate. Since point T is higher than point R, the y coordinate of point T must be greater than 1. Only answer choice D works.

The correct answer is D.
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by hey_thr67 » Mon Jun 04, 2012 9:06 am
Since we know that R is to be at the left of y=-x. We can infer that S is to be at the right of the y=-x and since, T has to be at the right of S, we can infer that T is in the first quadrant. Now our problems simplifies down whether T has Y- coordinate at 3 or 1. As Y coordinate of T can not be 1, we can easily check D as the answer.
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by anujan007 » Wed Aug 08, 2012 5:24 am
I had seen a similar problem before so it helped in the approach for this problem. Mitch's response was the best approach possible. Was able to nail down to D.

The approach to choose between D and E was particularly wonderful and a good learning.
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by rajeshsinghgmat » Wed May 08, 2013 6:20 pm
(D) (1, 3)
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by rdds » Thu Apr 07, 2016 3:13 am
Hi,
I understood the way to solve but the reason i didn't get to the solution is because I thought S could be in the 1st quadrant.(As shown in the image)
Image
Why is that not possible?
Also how to analyse the question so that I don't make such mistake.
Thanks in advance. :)
GMATGuruNY wrote:
arora007 wrote:In the rectangular coordinate system to the left, the line y = -x is the perpendicular bisector of segment RS (not shown) and the y-axis is the perpendicular bisector of segment ST (not shown). If the coordinates of point R are (-3, 1), what are the coordinates of point T ?

A (-1, 3)
B (-1, -3)
C (3, -1)
D (1, 3)
E (3, 1)

OA D

please help me with an easy explanation.
Also request you to help me with the simple approach to such problems.
Just draw a picture of what is being described. A perpendicular bisector:

-- intersects at the midpoint
-- forms a right angle

Here's my rudimentary drawing:

Image

Looking at the drawing above, we can see that T must have a positive x coordinate and a positive y coordinate. Since point T is higher than point R, the y coordinate of point T must be greater than 1. Only answer choice D works.

The correct answer is D.
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by GMATGuruNY » Thu Apr 07, 2016 7:08 pm
rdds wrote:Hi,
I understood the way to solve but the reason i didn't get to the solution is because I thought S could be in the 1st quadrant.(As shown in the image)
Image
Why is that not possible?
Also how to analyse the question so that I don't make such mistake.
Thanks in advance. :)
The prompt indicates that y=-x is the PERPENDICULAR BISECTOR of RS.
To bisect is TO DIVIDE INTO TWO EQUAL PARTS.
Since y=-x divides RS into two equal parts, the distance between y=-x and R must be equal to the distance between y=-x and S.
Thus, it is not possible that S is in the first quadrant.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by Matt@VeritasPrep » Mon Apr 11, 2016 1:15 pm
This is a bit of a controversial problem, I'd say: I don't *think* the GMAT can use a term like "perpendicular bisector" without defining it, since perpendicular bisectors (and medians, and angle bisectors, and orthocenters, and centroids, and a host of other fixtures of introductory geometry) are outside of the scope of the problems released so far.
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