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In the xy -plane, the point (1, 2) is on line j, and the poi

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In the xy -plane, the point (1, 2) is on line j, and the poi [#permalink] New post 24 Feb 2017, 02:26
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In the xy -plane, the point (1, 2) is on line j, and the point (2, 1) is on line k. Each of the lines has a positive slope.

Quantity A
Quantity B
The slope of line j
The slope of line k


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.
[Reveal] Spoiler: OA

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Re: In the xy -plane, the point (1, 2) is on line j, and the poi [#permalink] New post 28 Feb 2017, 16:31
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Explanation

Let us assume the line equations in terms of slope

line j: \(y = m1 \times x + c1\) ....... has the point (1,2) on it.

line k: \(y = m2 \times x + c2\) ....... has the point (2,1) on it.

Here m1 and m2 are slopes of the line j and k respectively.

Now putting the values (1,2) and (2,1) respectively into the equations of line j and k... we have

\(m1 + c1 = 2\) and \(2 \times m2 + c2 = 1\)

Let m1 = 5 and m2 = 2 and solve for c1 and c2 from their respective equations

c1 = -3 and c2 = -3


Line j: \(y = 5 \times x -3\)
Line k: \(y = 2 \times x -3\)

Clearly slope m1 is greater.

Now let m1 = 2 and m2 = 5 and solve for c1 and c2 from their respective equations

c1 = 0 and c2 = -9

Line j: \(y = 2 \times x\)
Line k: \(y = 5 \times x -9\)

Clearly slope m2 is greater.

Hence option D is correct.
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Re: In the xy -plane, the point (1, 2) is on line j, and the poi [#permalink] New post 30 Jul 2018, 11:02
Carcass wrote:


In the xy -plane, the point (1, 2) is on line j, and the point (2, 1) is on line k. Each of the lines has a positive slope.

Quantity A
Quantity B
The slope of line j
The slope of line k


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.

Please can you throw more light on this solution? I drew the xy-graph and line J appeared to have a steeper slope hence making the slope larger, IMO. Why are we using the xy graph equation to solve this?
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Re: In the xy -plane, the point (1, 2) is on line j, and the poi [#permalink] New post 31 Jul 2018, 13:40
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Emike56 wrote:
Please can you throw more light on this solution? I drew the xy-graph and line J appeared to have a steeper slope hence making the slope larger, IMO. Why are we using the xy graph equation to solve this?


How did you draw the lines? Only one point on each line is given and nothing is mentioned about the slope. Essentially you can draw any line passing through either points as lonfg as the slope is positive!
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Re: In the xy -plane, the point (1, 2) is on line j, and the poi [#permalink] New post 10 Sep 2018, 10:10
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simply put, we are just given two points in the xy plane and are told that the lines have positive slopes. this means that they both represent increasing functions. nothing else is given. just from the info of two numbers being positive doesn't give any drawable conclusion as to their comparison.

so the answer is D.

had it said that the increasing rate of one function is greater or less than the other, then we could've drawn a conclusion.
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Re: In the xy -plane, the point (1, 2) is on line j, and the poi [#permalink] New post 11 Sep 2018, 11:11
Easier way would be of think two different situations,

one is both the lines are parallel to x axis ie Slope is 0

and other take any random any position where the slope will differ.

SO D :)
Re: In the xy -plane, the point (1, 2) is on line j, and the poi   [#permalink] 11 Sep 2018, 11:11
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