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Math (Pre-Calculus)

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Find an equation of the line containing the points (8,-6) and (8, -4)

so far i've tried y-y1=m(x+x1) but currently cannot get the answer....T-T

the answer is suppose to be x=8

plz show the work so i can study form it and do it myself the next time... thnx in advance

  • Math (Pre-Calculus) -

    did you not notice that the slope is undefined?
    (slope = (-4 + 6)/(8-8) = 2/0)

    so you have a vertical line.
    All vertical lines are of the form x = ? , where the ? is simply the x coordinate of any point on the line
    Thus x = 8

    btw, a horizontal line would be of the form
    y = ? , where ? is the y coordinate of any point on the line , (all the y's would be the same).

    If you really insist on using your method, you could go ...

    y - (-6) = (2/0)(x - 8)

    then 0(y+6) = 2(x-8)
    0 = x-8
    x = 8
    but that is rather silly

  • Math (Pre-Calculus) -

    slope = m = (Y2-Y1)/(X2-X1) = (-4-8)/(8-8)
    Oh my, we have a problem, the slope is undefined because of zero denominator.
    The line with infinite slope goes straight up, constant x

    so the solution is
    x = 8

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