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    Question:

    Find equation of the line joining following points:

    (-3,4)(-3,9)

    So I found gradient or rather I didnt - ifyou find the gradient is it not infinity? 5/0...

    If I found a normal number I'd then do:

    y - y1 = m(x - x1)

    But I cant get that with infinity.

    The answer in the book is:

    x + 3 = 0
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    plot the points on an xy plane
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    (Original post by Pheylan)
    plot the points on an xy plane
    Ok I see the gradient is zero but then how do you possibly get x + 3?
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    Is it even possible to use the y -y1 = m (x - x1) formula for that?

    You see, with the points (-3,4) and (-3,9), it should be fairly clear to you that the it would be a vertical line passing through x = -3, hence, the answer.
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    Vertical lines (x = constant) can't be put into the form

    y = mx+c

    The most general form of a line's equation is

    ax+by+c=0
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    The line is vertical. You can still provide an equation for it though, it just won't be in the form y=mx+c.
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    (Original post by Dededex)
    Ok I see the gradient is zero but then how do you possibly get x + 3?
    Have you done transforming graphs yet?

    You need to understand that the line x=0 have been shifted 3 units in the negative x-direction. When going that direction you have to ADD the units. Therefore x + 3 = 0

    (When going the positive x-direction, you need to SUBTRACT the units, so if the points were (3, 4) and (3, 9) then the graph would be x - 3 = 0)

    EDIT:

    I just realised the line is just x = -3 where they've brought over the 3. You don't need to do all that but it's good practice :P:
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    (Original post by Dededex)
    Ok I see the gradient is zero but then how do you possibly get x + 3?
    gradient is infinite, not zero
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    It's a straight line... try plotting it on a graph if you can't see it
    It goes through (-3) , so it's the line x=-3 (or x+3=0 )
    You don't have to do what you did above
 
 
 
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