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Point K = (A,0), Point G = (2A + 4, sqrt2A+ 9). Is the dista [#permalink ]
10 Jan 2013, 08:55
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Point K = (A,0), Point G = (2A + 4, sqrt2A+ 9). Is the distance between point K and G prime?

(1) A^2 – 5A – 6 = 0

(2) A > 2

Last edited by

Bunuel on 10 Jan 2013, 15:39, edited 1 time in total.

RENAMED THE TOPIC.

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Re: Point K = (A,0), Point G = (2A + 4, sqrt2A+ 9). Is the dista [#permalink ]
10 Jan 2013, 15:51
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Point K = (A,0), Point G = (2A + 4, sqrt2A+ 9). Is the distance between point K and G prime? The formula to calculate the distance between two points

(x_1,y_1) and

(x_2,y_2) is

d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} .

Hence, the distance between points G and K is

d=\sqrt{(2A+4-A)^2+(\sqrt{2A+9}-0)^2}=\sqrt{(A+5)^2}=|A+5| (1) A^2 – 5A – 6 = 0 -->

(A-6)(A+1)=0 -->

A=6 or

A=-1 -->

d=|A+5|=11=prime or

d=|A+5|=4\neq{prime} . Not sufficient.

(2) A > 2. If

A=3 , then

d=|A+5|=8\neq{prime} but if

A=6 , then

d=|A+5|=11=prime . Not sufficient.

(1)+(2) Since from (2) A > 2, then from (1)

A=6 , thus

d=|A+5|=11=prime . Sufficient.

Answer: C.

Hope it's clear.

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Re: Point K = (A,0), Point G = (2A + 4, sqrt2A+ 9). Is the dista [#permalink ]
11 Jan 2013, 18:31
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Bunuel wrote:

Point K = (A,0), Point G = (2A + 4, sqrt2A+ 9). Is the distance between point K and G prime? The formula to calculate the distance between two points (x_1,y_1) and (x_2,y_2) is d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} . Hence, the distance between points G and K is d=\sqrt{(2A+4-A)^2+(\sqrt{2A+9}-0)^2}=\sqrt{(A+5)^2}=|A+5| (1) A^2 – 5A – 6 = 0 --> (A-6)(A+1)=0 --> A=6 or A=-1 --> d=|A+5|=11=prime or d=|A+5|=4\neq{prime} . Not sufficient. (2) A > 2. If A=3 , then d=|A+5|=8\neq{prime} but if A=-2 , then d=|A+5|=3=prime . Not sufficient. (1)+(2) Since from (2) A > 2, then from (1) A=6 , thus d=|A+5|=11=prime . Sufficient. Answer: C. Hope it's clear.

Hi Bunuel,

(2) A > 2. If

A=3 , then

d=|A+5|=8\neq{prime} but if A=6 , then d=|A+5|=11=prime . Not sufficient.

Coz it is given A>2 negative numbers are not possible right?

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Re: Coordinate Geometry [#permalink ]
10 Jan 2013, 10:20

Just to be clear the whole y coordinate for point G "2A+9" is all under the square root sign

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Re: Point K = (A,0), Point G = (2A + 4, sqrt2A+ 9). Is the dista [#permalink ]
12 Jan 2013, 03:26
shanmugamgsn wrote:

Bunuel wrote:

Point K = (A,0), Point G = (2A + 4, sqrt2A+ 9). Is the distance between point K and G prime? The formula to calculate the distance between two points (x_1,y_1) and (x_2,y_2) is d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} . Hence, the distance between points G and K is d=\sqrt{(2A+4-A)^2+(\sqrt{2A+9}-0)^2}=\sqrt{(A+5)^2}=|A+5| (1) A^2 – 5A – 6 = 0 --> (A-6)(A+1)=0 --> A=6 or A=-1 --> d=|A+5|=11=prime or d=|A+5|=4\neq{prime} . Not sufficient. (2) A > 2. If A=3 , then d=|A+5|=8\neq{prime} but if A=-2 , then d=|A+5|=3=prime . Not sufficient. (1)+(2) Since from (2) A > 2, then from (1) A=6 , thus d=|A+5|=11=prime . Sufficient. Answer: C. Hope it's clear.

Hi Bunuel,

(2) A > 2. If

A=3 , then

d=|A+5|=8\neq{prime} but if A=6 , then d=|A+5|=11=prime . Not sufficient.

Coz it is given A>2 negative numbers are not possible right?

Right. It should be:

(2) A > 2. If

A=3 , then

d=|A+5|=8\neq{prime} but if

A=6 , then

d=|A+5|=11=prime . Not sufficient.

Typo edited. Thank you. +1.

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Re: Point K = (A,0), Point G = (2A + 4, sqrt2A+ 9). Is the dista [#permalink ]
24 Jan 2014, 09:30

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