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In the xy-plane, point P has coordinates (a, b) and point Q has coordi

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xmagedo
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In the xy-plane, point P has coordinates (a, b) and point Q has coordi [#permalink] In the xy-plane, point P has coordinates (a, b) and point Q has coordi   Updated on: 21 Oct 2019, 01:59
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In the xy-plane, point P has coordinates (a, b) and point Q has coordinates (c, d). What is the distance between P and Q ?

(1) b - d = 4
(2) a - c = 3


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I dont understand this type of question, I mean the xy plane ! so if there a formula to follow in case then someone help me out


thanks
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C

Originally posted by xmagedo on 29 Jul 2010, 23:33.
Last edited by Bunuel on 21 Oct 2019, 01:59, edited 2 times in total.
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Re: In the xy-plane, point P has coordinates (a, b) and point Q has coordi [#permalink] In the xy-plane, point P has coordinates (a, b) and point Q has coordi   29 Jul 2010, 23:47
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xmagedo
In the xy plane, point p has coordinates (a,b) and point Q has coordinate(c,d). What is the distance between point Q ?

1 b-d=4
2 a-c=3


I dont understand this type of question, I mean the xy plane ! so if there a formula to follow in case then someone help me out


thanks

Distance between point \(A(x_1,y_1)\) and \(B(x_2,y_2)\) can be found by the formula \(D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).

So the distance between point \(P(a,b)\) and point \(Q(c,d)\) would be \(D=\sqrt{(a-c)^2+(b-d)^2}\).

(1) b-d=4 --> don't know a-c. Not sufficient.
(2) a-c=3 --> don't know b-d. Not sufficient.

(1)+(2) We have all necessary info. Sufficient: \(D=\sqrt{(a-c)^2+(b-d)^2}=\sqrt{3^2+4^2}=5\).

Answer: C.

For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature).

Hope it helps.
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Re: In the xy-plane, point P has coordinates (a, b) and point Q has coordi [#permalink] In the xy-plane, point P has coordinates (a, b) and point Q has coordi   19 Jul 2022, 10:05
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Re: In the xy-plane, point P has coordinates (a, b) and point Q has coordi [#permalink]
19 Jul 2022, 10:05
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