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Want to check the answer to this (Pythagoras)

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the coordinates of the points lying on the y-axis, P(0, y), when the distance between points P and A(5, 4) is 13.

    There is a diagram. Point B(0, 4) is also given. You can probably imagine how it looks like.

    2. Relevant equations

    3. The attempt at a solution

    The hypotenuse is 13.
    169 = 25 + (y + 4)^2

    In the answer it says (y - 4)^2 which doesn't make sense to me because the length from the origin to P(0, y) is y. y can either be above or below the x-axis but in both cases the length of BP will still be (y + 4).

    Please explain to me why length of BP is (y - 4)
  2. jcsd
  3. Nov 28, 2011 #2


    Staff: Mentor

    No, this should be 169 = 25 + (y - 4)2.

    The distance formula, which comes from the Pythagorean Theorem, says that the distance from A(x1, y1) to B(x2, y2) is sqrt((x2 - x1)2 + (y2 - y1)2).
  4. Nov 28, 2011 #3


    User Avatar
    Science Advisor

    It isn't and that does not say it is. We can make the line segment from [itex](x_1, y_1)[/itex] to [itex](x_2, y_2)[/itex] the hypotenuse of a right triangle by taking the third point, and the right angle, at either [itex](x_1, y_2)[/itex] or [itex](x_2, y_1)[/itex]. The two legs then have lengths [itex]|x_2- x_1|[/itex] and [itex]|y_2- y_1|[/itex].

    Here, the two endpoints are (0, y) and (5, 4). You can make that the hypotenuse of a right triangle by taking the third vertex at eight (5, y) or (0, 4).

    If you choose to use (5, y), then the distance from (0, y) to (5, y) is |0- 5|= 5 and the distance from (5, 4) to (5, y) is |4- y|= |y- 4| whether y is above or below 4.

    If you choose to use (0, 4), then the distance from (0, y) to (0, 4) is |y- 4|= |y- 4| and the distance from (5, 4) to (0, 4) is |5- 0|= 5. Either way the legs of the right triangle has legs of length 5 and |y- 4|. Of course once you square, it doesn't matter whether it is |y- 4| or |4- y|: [itex]|y- 4|^2=|4- y|^2= (y- 4)^2[/itex]. The square of the hypotenuse is [itex]25+ (y- 4)^2[/itex]. Set that equal to the square of 13 and solve for y.
  5. Nov 28, 2011 #4
    Okay thanks.

    In the diagram they plotted P(0, y) below the x-axis and I forgot to use the distance with absolute value thing so I thought it was y + 4 which seemed intuitive.
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