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Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), an

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Joined: 02 Sep 2009
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Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), an  [#permalink]

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23 Oct 2018, 04:04
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45% (medium)

Question Stats:

59% (01:27) correct 41% (01:26) wrong based on 102 sessions

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Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), and R = (9, 0). Rank these three points from closest to the origin, (0, 0), to furthest from the origin.

A. P, Q, R
B. P, R, Q
C. Q, P, R
D. R, P, Q
E. R, Q, P

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Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), an  [#permalink]

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23 Oct 2018, 04:40
2
Bunuel wrote:
Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), and R = (9, 0). Rank these three points from closest to the origin, (0, 0), to furthest from the origin.

A. P, Q, R
B. P, R, Q
C. Q, P, R
D. R, P, Q
E. R, Q, P

Distance between any two points $$(x_1, y_1)$$and $$(x_2, y_2)$$ $$= √[(x_2-x_1)^2 + (y_2-y_1)^2]$$

Distance of P from Origin, OP $$= √(8^2+4^2 = √80$$

Distance of Q from Origin, OP $$= √(6^2+7^2 = √85$$

Distance of R from Origin, OP $$= √(9^2+0^2 = √81$$

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Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), an  [#permalink]

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23 Oct 2018, 06:44
GMATinsight wrote:
Bunuel wrote:
Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), and R = (9, 0). Rank these three points from closest to the origin, (0, 0), to furthest from the origin.

A. P, Q, R
B. P, R, Q
C. Q, P, R
D. R, P, Q
E. R, Q, P

Distance between any two points $$(x_1, y_1)$$and $$(x_2, y_2)$$ $$= √[(x_2-x_1)^2 + (y_2-y_1)^2]$$

Distance of P from Origin, OP $$= √(8^2+4^2 = √80$$

Distance of Q from Origin, OP $$= √(6^2+7^2 = √85$$

Distance of R from Origin, OP $$= √(9^2+0^2 = √81$$

GMATinsight but this formula $$√[(x_2-x_1)^2 + (y_2-y_1)^2]$$ includes $$x_1$$ and $$x_2$$ and $$y_1$$ and $$y_2$$, whereas we have only P = (8, 4) i.e. $$x_1$$and $$y_1$$ can you please explain this phenomen ?
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
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Re: Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), an  [#permalink]

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23 Oct 2018, 06:48
1
dave13 wrote:
GMATinsight wrote:
Bunuel wrote:
Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), and R = (9, 0). Rank these three points from closest to the origin, (0, 0), to furthest from the origin.

A. P, Q, R
B. P, R, Q
C. Q, P, R
D. R, P, Q
E. R, Q, P

Distance between any two points $$(x_1, y_1)$$and $$(x_2, y_2)$$ $$= √[(x_2-x_1)^2 + (y_2-y_1)^2]$$

Distance of P from Origin, OP $$= √(8^2+4^2 = √80$$

Distance of Q from Origin, OP $$= √(6^2+7^2 = √85$$

Distance of R from Origin, OP $$= √(9^2+0^2 = √81$$

GMATinsight but this formula $$√[(x_2-x_1)^2 + (y_2-y_1)^2]$$ includes $$x_1$$ and $$x_2$$ and $$y_1$$ and $$y_2$$, whereas we have only P = (8, 4) i.e. $$x_1$$and $$y_1$$ can you please explain this phenomen ?

dave13

When we calculate length of OP then
$$(x_1, y_1)$$ = $$(8, 5)$$ and
$$(x_2, y_2)$$ = $$(0, 0)$$

i.e. OP = $$= √[(8-0)^2+(4-0)^2] = √(8^2+4^2) =√80$$

I hope this helps
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Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
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Re: Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), an   [#permalink] 23 Oct 2018, 06:48
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