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In the coordinate plane, points (a, b) and (c, d) are equidi
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30 Sep 2018, 15:58
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In the coordinate plane, points (a, b) and (c, d) are equidistant from the origin. \(|a| > |c|\)
Quantity A
Quantity B
|b|
|d|
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
Re: In the coordinate plane, points (a, b) and (c, d) are equidi
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10 Dec 2018, 17:41
5
Answer: B What we have:
Distance from (a,b) to origin (0,0) = √((a-0)^2 + (b-0)^2) = √(a^2 + b^2) Distance from (c,d) to origin (0,0) = √((c-0)^2 + (d-0)^2) = √(c^2 + d^2) It is said that points (a, b) and (c, d) are equidistant from the origin. It means their distances from origin are equal, so: √(a^2 + b^2) = √(c^2 + d^2) , both sides to power2: a^2 + b^2 = c^2 + d^2
As it is said that |a| > |c| so we can conclude that |a|^2 > |c|^2 [They are both positive, thus when they are in the same power, their comparison result will be just the same as when the power is 1.] Now that we know a^2 > c^2 and a^2 + b^2 = c^2 + d^2, so we can construe that b^2 is less than d^2 and so |b| is less than |d| (b and d don’t necessarily have the same relation) B is bigger than A
Re: In the coordinate plane, points (a, b) and (c, d) are equidi
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10 Dec 2018, 09:50
1
ruposh6240 wrote:
since both are equidistance from origin both are 5 , so ab (4,3) and cd (3,4) and and B
But d is bigger than b..
I'm sure the answer is wrong. Let's do it another way. We should have sqrt(b^2 + a^2) = sqrt(c^2 + d^2). squaring both sides we get b^2 + a^2 = c^2 + d^2 Since abs(a) > abs(c) we must have abs(b) < abs(d)
Re: In the coordinate plane, points (a, b) and (c, d) are equidi
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24 Mar 2024, 08:03
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Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink]