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Math Help - pythagorean

  1. #1
    Member helloying's Avatar
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    pythagorean

    A line cuts the x-axis at A(4,0) and the y-axis at B(0,3). Find the perdpendicular distance from O to L where O is the orgin. I took by the pythagors theorem 4² = 2.5² + ans²
    2.5 is half of the lenght of the line. Am i correct to do it this way?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by helloying View Post
    A line cuts the x-axis at A(4,0) and the y-axis at B(0,3). Find the perdpendicular distance from O to L where O is the orgin. I took by the pythagors theorem 4² = 2.5² + ans²
    2.5 is half of the lenght of the line. Am i correct to do it this way?
    hi.. you divided the line segment into 2, say at C.. but are you sure that the segment OC is perpendicular to AB?
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  3. #3
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    Yes, indeed. Kalagota implied that the point you found the distance from to the origin may not be on the line that contains point (0, 0) and perpendicular to the line in question. Do you get it now?
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  4. #4
    MHF Contributor kalagota's Avatar
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    ok, here are hints.

    1. Given a line l, get a point C(a,b)..

    using the point-slope form, you will be able to solve that point C, (i mean the coordinates of C) by solving simultaneously the two equations derived using the points C and the slope of the line AB.. (note that line AB contains C)

    2. After solving for the coordinates of C, then you can now solve the distance from O to C.
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  5. #5
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    Quote Originally Posted by helloying View Post
    A line cuts the x-axis at A(4,0) and the y-axis at B(0,3). Find the perdpendicular distance from O to L where O is the orgin. I took by the pythagors theorem 4² = 2.5² + ans²
    2.5 is half of the lenght of the line. Am i correct to do it this way?
    One way of doing this is by geometry.

    Let P be the point on line AB where the line from O intersects AB.
    Let u = line segment OP.

    Right triangle BOA is a 3-4-5 triangle, or hypotanuse AB = 5 units long.
    ((That can be shown by the Pythagorean theorem.))

    OP is perpendicular to AB
    OB is perpendicular to OA
    Therefore angle BOP is congruient to angle BAO.
    ((That is, angles whose sides are perpendicular each to each are congruent.))

    In right triangle BPO,
    cos (angle BOP) = u/3

    In right triangle BOA,
    cos(angle BAO) = 4/5

    Since angle BOP = angle BAO, then,
    cos(angle BOP) = cos(angle BAO)
    u/3 = 4/5
    u = 3*4/5 = 12/5 units long. --------answer.

    ----------------------------------------
    Another way is by "distance of a point to a line"

    The "line" is AB.
    Its equation, using the slope-intercept form, or y = mx +b, is
    y = (-3/4)x +3
    since its slope is -3/4 and its y-intercept is 3.

    In Ax +By +C = 0 form, that is
    (-3/4)x -y +3 = 0

    The "point" is the origin (0,0).

    The formula for the distance of a point to a line is
    d = |[A(x1) +B(y1) +C] / [sqrt(A^2 +B^2)]|

    So, since (x1,y1) = (0,0) here,
    d = |[(-3/4)*0 +(1)*0 +3] / [sqrt((-3/4)^2 +1^2)]|
    d = |[3] / [sqrt(9/16 +1)]|
    d = |[3] / [sqrt(25/16)]|
    d = 3 / (5/4)
    d = 3 * (4/5)
    d = 12/5 units. -------------same as above.
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  6. #6
    Member helloying's Avatar
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    Quote Originally Posted by kalagota View Post
    ok, here are hints.

    1. Given a line l, get a point C(a,b)..

    using the point-slope form, you will be able to solve that point C, (i mean the coordinates of C) by solving simultaneously the two equations derived using the points C and the slope of the line AB.. (note that line AB contains C)

    2. After solving for the coordinates of C, then you can now solve the distance from O to C.
    How to find the 2 equation derived using points C and slope of line AB? and i dont know how to find pont C also. Can anyone help me?
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  7. #7
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    Quote Originally Posted by helloying View Post
    How to find the 2 equation derived using points C and slope of line AB? and i dont know how to find pont C also. Can anyone help me?
    You have two points, (4,0) and (0,3), that are on the line in question. Now, to find the equation of the line, we must first find the slope (denoted by m):

    m = \frac{3-0}{0-4} = -\frac{3}{4}

    Now, to find the equation of the line (let's name it x), plug in the slope and any of the given points in this equation:

    y - y_1 = m(x-x_1)

    I'll do this one for you:

    y - 0 = -\frac{3}{4}(x-4)

    y = -\frac{3}{4}x + 3

    Alternatively, you could have simply replaced in slope-intercept form:

    y = mx + b

    where b is the y-intercept. But I needed to show you the steps. You have the equation of the line x now. What's next? Find the equation of the line that is perpendicular to line x. We'll name it y.

    You have to know that, to get any line y that is perpendicular to line x, line y should have a slope such that if I multiply it with the slope of line x, I get -1.

    Now you know how to find the slope of the perpendicular line, and you know one point on that line (don't forget it passes through the origin). Find the equation like I did before.

    Once you have the two equations, solve them simultaneously. Find the intersection point of the two lines. Then you know what to do next.
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  8. #8
    MHF Contributor kalagota's Avatar
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    chops has given you all the necessary steps..
    just feel free to ask what you dont understand..
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