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pappueshwar Really wants to Beat The GMAT! Default Avatar
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x plane Post Mon Mar 12, 2012 10:10 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    If Line K in the XY-Plane has equation y=mx+b, where m and b are constants, what is the slope of K?

    1. K is parallel to the line with equation y=(1-m)x+(b+1)
    2. K intersects the line with equation y=2x+3 at the point (2,7)


    OA IS A

    Request to explain as to how B is not the correct option

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    Post Mon Mar 12, 2012 8:08 pm
    You only have 1 point for line K, you need 2 to be able to figure out the slope.

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    Post Mon Mar 12, 2012 8:28 pm
    pappueshwar wrote:
    If Line K in the XY-Plane has equation y=mx+b, where m and b are constants, what is the slope of K?

    1. K is parallel to the line with equation y=(1-m)x+(b+1)
    2. K intersects the line with equation y=2x+3 at the point (2,7)


    OA IS A

    Request to explain as to how B is not the correct option
    Question: If equation of line K is y = mx + b, then what is the slope of K?

    (1) K is parallel to the line with equation y = (1 - m)x + (b + 1)
    Slope of given line, y = (1 - m)x + (b + 1) is (1 - m)
    Since slope of parallel lines are equal, so 1 - m = m implies m = 1/2; SUFFICIENT.

    (2) K intersects the line with equation y = 2x + 3 at the point (2,7).
    Since line K passes through the point (2, 7), so this point will satisfy y = mx + b
    So, 7 = 2m + b but this does help us to find the value of m as there is just one equation and 2 variables; NOT sufficient.

    The correct answer is A.

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    Post Tue Mar 13, 2012 4:14 am
    If Line K in the XY-Plane has equation y=mx+b, where m and b are constants, what is the slope of K?

    1. K is parallel to the line with equation y=(1-m)x+(b+1)
    2. K intersects the line with equation y=2x+3 at the point (2,7)


    I got this answer as wrong... Could any of you tell me what is the difference between the y intercept i.e b in this case and the 'y' itself i.e 7 in this case...

    This confusion has been in my mind since ages...

    I really need an answer to this anyhow...

    Thanking You...

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