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# What is the equation that represents the locus of points equidistant

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Joined: 18 May 2008
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What is the equation that represents the locus of points equidistant  [#permalink]

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24 Jun 2008, 03:20
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55% (02:15) correct 45% (02:07) wrong based on 85 sessions

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What is the equation that represents the locus of points equidistant from the points A (-2, 1) and B (6, 3)?

(A) 4x+y=20
(B) y+4x=10
(C) 4y+x=10
(D) y+2x=20
(E) x+4y=40
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Joined: 12 Jun 2008
Posts: 286
Schools: INSEAD Class of July '10
Re: What is the equation that represents the locus of points equidistant  [#permalink]

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24 Jun 2008, 03:35

The locus of points equidistant from 2 points is the perpendicular bisector of the segment between these 2 points.

Vector AB is (8,2) and we want a vector (x,y) which is normal to that one (it will give the direction of the perpendicular bisector).

Their dot product has to be null, so we get 8x+2y=0, which is equal to 4x+y=0

Therefore any line with equation 4x+y=K is perpendicular to the line that goes through A and B.

Answer is then either (A) or (B).

If the line we are looking for is the locus of points equidistant from A and B, it contains the middle point between A and B, which is M(2,2).

This point verifies 4x+y=10: (B) is the answer.
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Re: What is the equation that represents the locus of points equidistant  [#permalink]

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24 Jun 2008, 05:39
ritula wrote:
What is the equation that represents the locus of points equidistant from the points A (-2, 1) and B (6, 3)?
(A) 4x+y=20
(B) y+4x=10
(C) 4y+x=10
(D) y+2x=20
(E) x+4y=40
I know that its pretty simple question, but sumhow im not getting the right answer. pls help!

B

[x-(-2)]^2 + [y -1]^2 = [x-6]^2 + [y-3]^2

the squared terms on both sides get cancelled leaving us with B
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Re: What is the equation that represents the locus of points equidistant  [#permalink]

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24 Jun 2008, 05:56
1
Maybe I just like to work harder than everyone else. Here is how I arrived at B.

Locus = perpendicular bisector.

Find the mid point between A & B. 6 - -2 = 8 and 3 - 1 = 2, so go right 8 and up 2, so midpoint is half that. Go right 4, and up 1. Brings us to {2, 2}. Slope of AB is 1/4 because rise over run 2 up 8 right = $$\frac{2}{8}$$ or $$\frac{1}{4}$$.

The prependicular line to this will be inverted and the sign flipped. Inverse of 1/4 is 4/1 and negative is -4. To find the x-intercept you have to move left from 2,2 so go up 4, but left, or negative 1. (If you have 4/1, if you move in a positive direction for one of the numbers 4 or 1, the other must move in a negative direction). Up 4 and left 1 brings you to point 1, 6, that's not the x yet, so up 4 and left 1 again brings us to 0 ,10. x-intercept of +10.

Now this perpendicular line is represented by the equation y = -4x + 10. Find the answer that will give you that when solved for y. This is B

y + 4x = 10
y = 10 - 4x or
y = -4x +10

ritula wrote:
What is the equation that represents the locus of points equidistant from the points A (-2, 1) and B (6, 3)?
(A) 4x+y=20
(B) y+4x=10
(C) 4y+x=10
(D) y+2x=20
(E) x+4y=40
I know that its pretty simple question, but sumhow im not getting the right answer. pls help!

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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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Joined: 18 May 2008
Posts: 1134
Re: What is the equation that represents the locus of points equidistant  [#permalink]

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24 Jun 2008, 23:00
OA is B indeed. i was doing the big mistake of keeping its slope the same as the given line while actually it is a perpendicular line. Thanks to all!
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Joined: 23 Nov 2014
Posts: 60
What is the equation that represents the locus of points equidistant f  [#permalink]

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12 May 2016, 18:49
What is the equation that represents the locus of points equidistant from the points A (-2,1) and B (6,3)

A) 4x+y = 20
B) y+4x = 10
C) 4y+x = 10
D) y+2x = 20
E) x+4y = 20

My Solution is to find the slope of line AB, which is equal to 1/4. The equation of locus of points asked thus will have a slope of -4.

That itself allows me to reject C,D and E
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Joined: 02 Aug 2009
Posts: 6956
Re: What is the equation that represents the locus of points equidistant f  [#permalink]

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12 May 2016, 20:40
2
2
ruchi857 wrote:
What is the equation that represents the locus of points equidistant from the points A (-2,1) and B (6,3)

A) 4x+y = 20
B) y+4x = 10
C) 4y+x = 10
D) y+2x = 20
E) x+4y = 20

My Solution is to find the slope of line AB, which is equal to 1/4. The equation of locus of points asked thus will have a slope of -4.

That itself allows me to reject C,D and E

hi,

1) A (-2,1) and B (6,3)..
therefore a point equidistant from BOTH will be there average..
so$$x = \frac{-2+6}{2}= 2$$ and $$y =\frac{1+3}{2} = 2$$..
substitute x as 2 and y as 2 and see which equation satisfies the values...
ONLY B and C are left..

2) As shown by you, work on slope..
slope of line AB =$$\frac{3-1}{6-(-2)} = \frac{2}{8} = \frac{1}{4}$$..
so slope of line joining all points = -4, as this line will be perpendicular to AB...
ONLY A and B satisfy the condition

ONLY B satisfies both points..
ans B
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Re: What is the equation that represents the locus of points equidistant  [#permalink]

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13 May 2016, 00:17
ruchi857 wrote:
What is the equation that represents the locus of points equidistant from the points A (-2,1) and B (6,3)

A) 4x+y = 20
B) y+4x = 10
C) 4y+x = 10
D) y+2x = 20
E) x+4y = 20

My Solution is to find the slope of line AB, which is equal to 1/4. The equation of locus of points asked thus will have a slope of -4.

That itself allows me to reject C,D and E

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Re: What is the equation that represents the locus of points equidistant  [#permalink]

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24 Sep 2018, 07:44
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Re: What is the equation that represents the locus of points equidistant &nbs [#permalink] 24 Sep 2018, 07:44
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