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16 Sep 2014, 01:16
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In the coordinate plane, is the point \((0, 0)\) closer to the point \((u, v)\) than to the point \((u, v + 1)\)?
(1) \(v + u^2 = -1\)
(2) \(v < 0\)
(1) \(v + u^2 = -1\)
(2) \(v < 0\)
16 Sep 2014, 01:16
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Official Solution:
In the coordinate plane, is the point \((0, 0)\) closer to the point \((u, v)\) than to the point \((u, v + 1)\)?
The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).
Essentially, the question asks whether the distance between the points \((0, 0)\) and \((u, v)\) is less than the distance between the points \((0, 0)\) and \((u, v + 1)\):
Is \(\sqrt{(u-0)^2+(v-0)^2} \lt \sqrt{(u-0)^2+(v+1-0)^2}\)?
Is \(\sqrt{u^2+v^2} \lt \sqrt{u^2+(v+1)^2}\)?
Is \(u^2+v^2 \lt u^2+v^2+2v+1\)?
Is \(v \gt -\frac{1}{2}\)?
(1) \(v + u^2 = -1\).
Rearrange: \(v=-1-u^2\). Now, since \(u^2 \ge 0 \), then \(v=-1-u^2 \le -1\), so the answer to the question is NO. Sufficient.
(2) \(v < 0\). Not sufficient.
Answer: A
In the coordinate plane, is the point \((0, 0)\) closer to the point \((u, v)\) than to the point \((u, v + 1)\)?
The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).
Essentially, the question asks whether the distance between the points \((0, 0)\) and \((u, v)\) is less than the distance between the points \((0, 0)\) and \((u, v + 1)\):
Is \(\sqrt{(u-0)^2+(v-0)^2} \lt \sqrt{(u-0)^2+(v+1-0)^2}\)?
Is \(\sqrt{u^2+v^2} \lt \sqrt{u^2+(v+1)^2}\)?
Is \(u^2+v^2 \lt u^2+v^2+2v+1\)?
Is \(v \gt -\frac{1}{2}\)?
(1) \(v + u^2 = -1\).
Rearrange: \(v=-1-u^2\). Now, since \(u^2 \ge 0 \), then \(v=-1-u^2 \le -1\), so the answer to the question is NO. Sufficient.
(2) \(v < 0\). Not sufficient.
Answer: A
15 Jul 2020, 04:56
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I think this is a high-quality question and I agree with explanation.
