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In the rectangular coordinate system, (r, s) and (u, v) are two points
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13 Dec 2023, 01:01
13 Dec 2023, 01:01
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In the rectangular coordinate system, (r, s) and (u, v) are two points. It is known that r = v =1 − s and u = 1 − r .
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.
Quantity A |
Quantity B |
Distance of the point (r, s) from the origin |
Distance of the point (u, v) from the origin |
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.
Part of the project: The Butler-GRE Daily New Quant and Verbal Questions to Practice (2023) - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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In the rectangular coordinate system, (r, s) and (u, v) are two points
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17 Jan 2024, 13:14
17 Jan 2024, 13:14
1
The distance to origin formula is \(d = \sqrt{x^2+y^2}\)
Distance of point (r, s) from origin is
\(A = \sqrt{r^2+s^2}\)
Distance of point (u, v) from origin is
\(B = \sqrt{u^2+v^2}\)
Since \(r = v\), \(r^2 = v^2\)
Since \(r = 1 - s\), \(s = r - 1\) → \(s^2 = (1 - r)^2\)
Since \(u = 1 - r\), \(u^2 = (1 - r)^2\)
\(A = \sqrt{r^2+s^2} = \sqrt{r^2+(1 - r)^2}\)
\(B = \sqrt{u^2+v^2} = \sqrt{(1 - r)^2+r^2}\)
A = B
Distance of point (r, s) from origin is
\(A = \sqrt{r^2+s^2}\)
Distance of point (u, v) from origin is
\(B = \sqrt{u^2+v^2}\)
Since \(r = v\), \(r^2 = v^2\)
Since \(r = 1 - s\), \(s = r - 1\) → \(s^2 = (1 - r)^2\)
Since \(u = 1 - r\), \(u^2 = (1 - r)^2\)
\(A = \sqrt{r^2+s^2} = \sqrt{r^2+(1 - r)^2}\)
\(B = \sqrt{u^2+v^2} = \sqrt{(1 - r)^2+r^2}\)
A = B
In the rectangular coordinate system, (r, s) and (u, v) are two points
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23 Jan 2024, 19:45
23 Jan 2024, 19:45
1
The distance between any two points, \((x1,y1)\) and \((x2,y2)\), is given by the Pythagoras Theorem
\(d = \sqrt{(x2-x1)^2 + (y2-y1)^2}\)
Since in this case \((x1,y1) = (0,0)\)
\(d = \sqrt{x^2 + y^2}\)
Quantity A
Distance of the point \((r,s)\) from the origin
\(d = \sqrt{r^2 + s^2}\)
\(d= \sqrt{r^2 +(1 - r)^2}\)
\(d=\sqrt{r^2 + 1^2 + r^2 - 2r}\)
\(d= \sqrt{2r^2 -2r + 1}\)
Quantity B
Distance of point \((u,v)\) from the origin
\(d = \sqrt{u^2 + v^2}\)
Since \(u = 1-r\) and \(v=r=1-s \)
\(d= \sqrt{(1-r)^2 + r^2}\)
\(d = \sqrt{1^2 + r^2 -2r + r^2}\)
\(d = \sqrt{2r^2-2r+1}\)
BOTH quantities are EQUAL
The answer is C
\(d = \sqrt{(x2-x1)^2 + (y2-y1)^2}\)
Since in this case \((x1,y1) = (0,0)\)
\(d = \sqrt{x^2 + y^2}\)
Quantity A
Distance of the point \((r,s)\) from the origin
\(d = \sqrt{r^2 + s^2}\)
\(d= \sqrt{r^2 +(1 - r)^2}\)
\(d=\sqrt{r^2 + 1^2 + r^2 - 2r}\)
\(d= \sqrt{2r^2 -2r + 1}\)
Quantity B
Distance of point \((u,v)\) from the origin
\(d = \sqrt{u^2 + v^2}\)
Since \(u = 1-r\) and \(v=r=1-s \)
\(d= \sqrt{(1-r)^2 + r^2}\)
\(d = \sqrt{1^2 + r^2 -2r + r^2}\)
\(d = \sqrt{2r^2-2r+1}\)
BOTH quantities are EQUAL
The answer is C
