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M12-13

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Re: M12-13  [#permalink]

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New post 13 Sep 2017, 23:32
d975490 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. i read the whole thread, and still didn't understand the explanation. I do understand that we don't need to find the value of a and b, question is why would we discard 1) and 2) as insufficient, what's the reasoning behind it.


(1) and (2) individually are insufficient for the same reason: different values of a and b satisfy each statement, giving different points of (a, b), thus giving different equations of line k that is perpendicular to line y = 2x.

For example, for (1), if x = 2 and y = -2, then the equation of line k would be y = -x/2 - 1 (apply the approach used HERE) but of x = 1 and y = -1, then the equation of line k would be y = -x/2 - 1/2. Two different answers, hence insufficient.

Hope it's clear.
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Re: M12-13  [#permalink]

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New post 31 Dec 2017, 12:55
Bunuel wrote:
Official Solution:


Notice that we can get the equation of line \(k\) which is perpendicular to line \(y=2x\) if we know ANY point that line \(k\) passes through. So, to get the equation of line \(k\) we need the values of \(a\) and \(b\).

(1) \(a = -b\). Not sufficient.

(2) \(a - b = 1\). Not sufficient.

(1)+(2) \(a = -b\) and \(a - b = 1\), we have two distinct linear equations with two unknowns so we can solve for \(a\) and \(b\). Sufficient.


Answer: C


Hi, Bunuel!

Is it necessary to find the equation of line in absolute values of 'a' and 'b'? I mean the point (a,b) is given. Can't we just take 'a' and 'b' as known values because point (a,b) is mentioned?
My first thought while solving this question was that slope is given and point (a,b) is given through which the line passes. Only 'c' is unknown to find the equation of line. Taking each of the statements one by one, i was able to find the value of 'c' in terms of 'a' or 'b' and hence the equation of line in terms of all the known values i.e 'a' and 'b' was obtained.
So, this way each statement was sufficient according to me and I thought D to be the answer.

Please, clarify my this doubt.Thank you.
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Re: M12-13  [#permalink]

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New post 06 Jul 2018, 14:25
Bunuel wrote:
subhajit1 wrote:
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

So, statement 1 seems sufficient (contradiction with the actual answer)


Check the diagram below:

Image

As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.

Attachment:
Untitled.png



Sorry to be dense about this - if the line passes through points (1,-1); (0,0) and (2,-2), then we know both the slope and the x intercept. So we know the equation of the line?
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Re: M12-13  [#permalink]

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New post 07 Jul 2018, 00:37
londonconsultant wrote:
Bunuel wrote:
subhajit1 wrote:
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

So, statement 1 seems sufficient (contradiction with the actual answer)


Check the diagram below:

Image

As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.

Attachment:
Untitled.png



Sorry to be dense about this - if the line passes through points (1,-1); (0,0) and (2,-2), then we know both the slope and the x intercept. So we know the equation of the line?


We want to find the equation of line k. The diagram there shows TWO possible lines for k: RED and GREEN. They have different equations. So, we cannot get one single equation to answer the question.
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M12-13  [#permalink]

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New post 07 Jul 2018, 06:43
But the lines do not satisfy b = -a. For example, where the lines cross the x axis should be (0,0), but we can see these lines have x values of -2 and -1 respectively

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New post 07 Jul 2018, 08:54
londonconsultant wrote:
But the lines do not satisfy b = -a. For example, where the lines cross the x axis should be (0,0), but we can see these lines have x values of -2 and -1 respectively

Bunuel


Not sure what you are trying to say there.

We are told that k passes through the point (a, b). (1) says that a = -b. The image above shows two possible cases (out of infinitely many) for the point (a, b):
If a = 1 and b = -1, then the equation of k is -x/2 - 1/2.
If a = 2 and b = -2, then the equation of k is -x/2 - 1.
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Re: M12-13 &nbs [#permalink] 07 Jul 2018, 08:54

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