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In the coordinate plane above, point C is not displ

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In the coordinate plane above, point C is not displ [#permalink] New post 02 Aug 2017, 08:12
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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)
[Reveal] Spoiler: OA

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Re: In the coordinate plane above, point C is not displ [#permalink] New post 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] New post 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] New post 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: (x-h)^2 + (y-k)^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.

(x-3)^2 + (y-4)^2 = 10^2

Finally, we plug in the x and y coordinates of the answer choices in the formula.

A) (-5-3)^2 + (-2-4)^2 = 10^2
64 + 36 =100

...

C) (10-3)^2 + (11-4)^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] New post 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] New post 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: (x-h)^2 + (y-k)^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.

(x-3)^2 + (y-4)^2 = 10^2

Finally, we plug in the x and y coordinates of the answer choices in the formula.

A) (-5-3)^2 + (-2-4)^2 = 10^2
64 + 36 =100

...

C) (10-3)^2 + (11-4)^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
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In the coordinate plane above, point C is not displ

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