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In the coordinate plane above, point C is not displ [#permalink]
02 Aug 2017, 08:12
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#GREpracticequestion In the coordinate plane above, point C is not displayed.jpg [ 9.24 KiB  Viewed 1603 times ]
In the coordinate plane above, point C is not displayed. If the length of line segment BC is twice the length of line segment AB, which of the following could not be the coordinates of point C? (A) (5,2) (B) (9, 12) (C) (10, 11) (D) (11, 10) (E) (13, 4)
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Re: In the coordinate plane above, point C is not displ [#permalink]
19 Sep 2017, 07:48
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Since the AB segment has one extreme in the origin (0,0), its length can be measures as \(sqrt(3^2+4^2)=sqrt(25)=5\). The segment BC = 2AB is thus 10 long. Then, given the formula for the distance between two points we have to look for the couplet whose distance from B is different from 10. Since choice C implies a distance of \(4sqrt(6)<10\), this is the answer!



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Re: In the coordinate plane above, point C is not displ [#permalink]
18 Sep 2018, 14:10
In order to solve this, find AB= 5, I.E BC= 10 BC^2=100 Now using backsolving, see which coordinates do not give 100 as the squared difference of B and C. The answer is C.



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Re: In the coordinate plane above, point C is not displ [#permalink]
13 Jan 2019, 05:00
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AB is the hypotenuse of the triangle which has vertices of (0/0), (3/0) and (3/4). The lengths of the two shorter sides are 3 (horizontal side) and 4 (vertical side). This is the most common form of a 3  4  5 Pythagorean triple. Thus the hypotenuse is has a length of 5.
Twice the line segment equals a distance of 10. So every point which has a distance equal to 10 to point B is a possible coordinate. Since the point can be anywhere with a distance of 10, we can think of the set of possible places for point C as all points on the "outer line of" a circle with a radius of 10.
Circle formula: (xh)^2 + (yk)^2 = r^2 ,where (h/k) is the center of the circle and r is the radius. In this case, the center is (3/4) and the radius is 10.
Thus we plug in the given values in the formula.
(x3)^2 + (y4)^2 = 10^2
Finally, we plug in the x and y coordinates of the answer choices in the formula.
A) (53)^2 + (24)^2 = 10^2 64 + 36 =100
...
C) (103)^2 + (114)^2 = 10^2 49 + 49 =/= 100 Here, the equation is not fulfilled and thus the distance from B to C is shorter than 10.



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Re: In the coordinate plane above, point C is not displ [#permalink]
05 Sep 2019, 17:55
Any other solutions for this question?



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Re: In the coordinate plane above, point C is not displ [#permalink]
03 May 2020, 05:30
Zamala wrote: AB is the hypotenuse of the triangle which has vertices of (0/0), (3/0) and (3/4). The lengths of the two shorter sides are 3 (horizontal side) and 4 (vertical side). This is the most common form of a 3  4  5 Pythagorean triple. Thus the hypotenuse is has a length of 5.
Twice the line segment equals a distance of 10. So every point which has a distance equal to 10 to point B is a possible coordinate. Since the point can be anywhere with a distance of 10, we can think of the set of possible places for point C as all points on the "outer line of" a circle with a radius of 10.
Circle formula: (xh)^2 + (yk)^2 = r^2 ,where (h/k) is the center of the circle and r is the radius. In this case, the center is (3/4) and the radius is 10.
Thus we plug in the given values in the formula.
(x3)^2 + (y4)^2 = 10^2
Finally, we plug in the x and y coordinates of the answer choices in the formula.
A) (53)^2 + (24)^2 = 10^2 64 + 36 =100
...
C) (103)^2 + (114)^2 = 10^2 49 + 49 =/= 100 Here, the equation is not fulfilled and thus the distance from B to C is shorter than 10. THanks, this solution is by far the best way to see it I believe




Re: In the coordinate plane above, point C is not displ
[#permalink]
03 May 2020, 05:30





