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Math Help - Show that gradient is m for all positions...

  1. #1
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    Show that gradient is m for all positions...

    There are two questions that I have no idea how to solve:

    1. Let P, with coordinates(p,q) be a fixed point on the curve with equation y = mx + c and let Q, with coordinates (r,s), be any other point on y = mx + c. Use the fact that the coordinates of P and Q satisfy the equation y = mx + c to show that the gradient of the PQ is m for all positions of Q.

    2. There are some values of a, b, and c for which the equation ax + by + c = 0 does not represent a straight line. Give an example of such values.
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  2. #2
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    Quote Originally Posted by struck View Post
    There are two questions that I have no idea how to solve:

    1. Let P, with coordinates(p,q) be a fixed point on the curve with equation y = mx + c and let Q, with coordinates (r,s), be any other point on y = mx + c. Use the fact that the coordinates of P and Q satisfy the equation y = mx + c to show that the gradient of the PQ is m for all positions of Q.

    2. There are some values of a, b, and c for which the equation ax + by + c = 0 does not represent a straight line. Give an example of such values.
    I cannot quite understand the first question. Do you mean point Q is a random point? Q is at (r,s) now, but Q may be at (t,u), or (v,w), or (x,y), or.....?

    For the fixed point P,
    q = m(p) +c --------------(1)

    For the random point Q, which is at (r,s) now,
    s = m(r) +c -------------(2)

    Eq.(1) minus Eq.(2),
    q -s = m(p) -m(r)
    q -s = m(p -r)
    m = (q-s) /(p -r) ------(i)**

    If Q is at another position, say at (x,y),
    y = m(x) +c ------------(3)

    Eq.(1) minus Eq.(3),
    q -y = m(p) -m(x)
    q -y = m(p -x)
    m = (q -y) /(p -x) ------(ii)**

    Now, compare Eq.(i) and Eq.(ii).
    The gradient m remains the same even if point Q transferred from (r,s) to (x,y).

    Is that what your first question wants to see?

    -------------------------------------------------------
    Your second question.

    ax +by +c = 0 --------(4)

    Eq.(4) will always be a straight line as long as there is a variable ...x or y... left on the equation and a non-xero c is also left in the equation. Meaning,
    ax +c = 0 is a line. ----**
    by +c = 0 is a line. ----**
    c = 0 is not a line.

    ax = 0 is not a line
    by = 0 is not a line.
    However, ax +by = 0 is a line. ----**

    Therefore, if any two of the a,b,c are zeros at the same time, there will be no straight line.
    So, a=0, b=0, c = any number -----no straight line.
    a = 0, b = any number, c = 0 -------no line.
    a = any number, b=0, c=0 ----------no line.
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