Hello,
I was wondering if you can please help with the explanation here. This is from OG 13 P. 318
74) In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3)?
1) The x-coordinate of point R is -1
2) Point R lies on the line y = -3
[spoiler]OA: A[/spoiler]
Thanks a lot.
Best Regards,
Sri
Coordinate Geometry
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we need to use a distance formula here for point R (x,y): (x-(-3))^2+(y-(-3))^2=(x-1)^2+(y-(-3))^2 or (x+3)^2+(y+3)^2=(x-1)^2+(y+3)^2 or as evident from the left-hand side and right-hand side we have the same (y+3)^2, therefore, (x+3)^2=(x-1)^2 <-> 8x=-8 and x=-1gmattesttaker2 wrote:Hello,
I was wondering if you can please help with the explanation here. This is from OG 13 P. 318
74) In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3)?
1) The x-coordinate of point R is -1
2) Point R lies on the line y = -3
[spoiler]OA: A[/spoiler]
Thanks a lot.
Best Regards,
Sri
Now, if we encounter in any statement x=-1 we should get sufficient information about our point R with coordinates (x,y) as y is not considered in our distance relationship at all.
st(1) x=-1 ==> Sufficient
st(2) y=-3 but we don't know about x-coordinate, Not Sufficient
answer a
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Hi!gmattesttaker2 wrote:Hello,
I was wondering if you can please help with the explanation here. This is from OG 13 P. 318
74) In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3)?
1) The x-coordinate of point R is -1
2) Point R lies on the line y = -3
[spoiler]OA: A[/spoiler]
This question should take about 15 seconds if you actually draw out the x-y plane. There are no bonus points on the GMAT for super-complicated math, so avoid it whenever possible!
If you plot the 2 points, you can see that, since they have the same y-coordinate (-3), one point that's halfway between them is (-1, -3) (the difference in x-coordinates is 4, so if you travel 2 units from either point you'll end up in the middle).
Further, if you draw a vertical line through x=-1, you can see that every point on that line is equidistant from the two points given.
Accordingly, if we know that point R has an x-coordinate of -1, the answer will be YES; if point R has any other x-coordinate, the answer will be NO.
1) gives us the x-coordinate... sufficient!
2) doesn't give us the x-coordinate... insufficient!
(1) is suff, (2) isn't... choose A!
Moral of the story: on geometry word problems, always draw out what you know.
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we got lucky that two points of the line segment were given with the same y-coordinate, i.e. parallel to x-abscess. What if we have different points?Stuart Kovinsky wrote:There are no bonus points on the GMAT for super-complicated math, so avoid it whenever possible!
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We didn't get lucky - the question was designed to reward people who saw the quick solution.pemdas wrote:we got lucky that two points of the line segment were given with the same y-coordinate, i.e. parallel to x-abscess. What if we have different points?Stuart Kovinsky wrote:There are no bonus points on the GMAT for super-complicated math, so avoid it whenever possible!
Most GMAT math questions are designed with 4 levels of reward:
1) negative reward, for people who spend time on it and get it wrong;
2) 0 reward, for people who recognize that they don't know how to solve and get it wrong, but don't waste time on the question;
3) positive reward, for people who take traditional approaches to questions - they get it right, but they spend a lot of time doing so; and
4) super positive reward, for people who find the most efficient solution to the problem - they get it right in a lot less time, leaving more time for other questions.
Never forget - while the GMAT tests math, it's not a math test; the primary skills tested on the GMAT are problem analysis and strategic thinking.
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We need to determine whether point R is equidistant from points (-3,-3) and (1,-3). We know that if point R is on the perpendicular bisector of the line segment with (-3,-3) and (1,-3) as the endpoints, then R is equidistant from (-3,-3) and (1,-3). Since the midpoint of the line segment with endpoints (-3,-3) and (1,-3) is (-1, -3) and the line segment is horizontal, the perpendicular bisector is a vertical line passing through (-1, -3). That is, the perpendicular bisector has equation x = -1. Thus, if point R has an x-coordinate of -1, then point R will be equidistant from (-3,-3) and (1,-3) .gmattesttaker2 wrote:Hello,
I was wondering if you can please help with the explanation here. This is from OG 13 P. 318
74) In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3)?
1) The x-coordinate of point R is -1
2) Point R lies on the line y = -3
Statement One Alone:
The x-coordinate of point R is -1.
Since the x-coordinate of point R is -1, R is on the line with equation x = -1. That is, R is on the perpendicular bisector of the line segment with endpoints (-3,-3) and (1,-3), so R is equidistant from (-3,-3) and (1,-3). Statement one alone is sufficient to answer the question.
Statement Two Alone:
Point R lies on the line y = -3.
Statement two is not sufficient to answer the question. For example, if R = (-1, -3), then it's equidistant from (-3,-3) and (1,-3). However, if R = (0, -3), then it's not equidistant from (-3,-3) and (1,-3).
Answer: A
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