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    Find the Cartesian equation of the set of point P such that P is at a constant distance of
    five units from the line 4x−3y =1.

    I don't understand how I am supposed to go about answering this question :confused:

    Thanks in advance!
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    (Original post by YdLeet)
    Find the Cartesian equation of the set of point P such that P is at a constant distance of
    five units from the line 4x−3y =1.

    I don't understand how I am supposed to go about answering this question :confused:

    Thanks in advance!
    First you would to know that when a given P_0(x_0,y_0) is on a line and \vec n(A,B) is the normal vector being perpendicular to that line and let the
    P(x,y) be a running point on the line then the equation of the line is
    \vec P_0P \cdot \vec n=0
    that is the dot product of vector P_0P and the normal vector is zero because they are perpendicular to each other.

    With coordinates
    A(x-x_0)+B(y-y_0)=0
    Arranging the equation
    Ax+By=Ax_0+By_0
    At RHS there is a constant.
    In your example this is 1, and n(4, -3)
    For all P(x,y) which the equation is true are on the line.
    In your example fe. P(1,1)
    4x-3y-1=0
    so this point is on this line.

    When the P in not on the line then the equation gives a number, the dot product of PoP and the normal vector (which won't be zero), and this value will be the projection of PoP on the normal vector.
    If the magnitude of the normal vector would be 1, this value would give the distance of P from the line, followed from the definition of the dot product.
    \vec P_0P\cdot \vec n=|\vec P_0P|\cdot |\vec n|\cdot cos \gamma

    The magnitude of the normal vector is
    |\vec n|=\sqrt{A^2+B^2}
    So the coordinates of unit vector in direction of the normal vector are
    \displaystyle \vec e_n\left (\frac{A}{\sqrt{A^2+B^2}},\frac{  B}{\sqrt{A^2+B^2}} \left )
    With this vector the equation of the line gives zero when P(x,y) on the line
    and the signed distance of P from the line if it is not on the line.
    So this distance
    \displaystyle \left |\frac{Ax+By+C}{\sqrt{A^2+B^2}} \right |=d
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    (Original post by ztibor)
    First you would to know that when a given P_0(x_0,y_0) is on a line and \vec n(A,B) is the normal vector being perpendicular to that line and let the
    P(x,y) be a running point on the line then the equation of the line is
    \vec P_0P \cdot \vec n=0
    that is the dot product of vector P_0P and the normal vector is zero because they are perpendicular to each other.

    With coordinates
    A(x-x_0)+B(y-y_0)=0
    Arranging the equation
    Ax+By=Ax_0+By_0
    At RHS there is a constant.
    In your example this is 1, and n(4, -3)
    For all P(x,y) which the equation is true are on the line.
    In your example fe. P(1,1)
    4x-3y-1=0
    so this point is on this line.

    When the P in not on the line then the equation gives a number, the dot product of PoP and the normal vector (which won't be zero), and this value will be the projection of PoP on the normal vector.
    If the magnitude of the normal vector would be 1, this value would give the distance of P from the line, followed from the definition of the dot product.
    \vec P_0P\cdot \vec n=|\vec P_0P|\cdot |\vec n|\cdot cos \gamma

    The magnitude of the normal vector is
    |\vec n|=\sqrt{A^2+B^2}
    So the coordinates of unit vector in direction of the normal vector are
    \displaystyle \vec e_n\left (\frac{A}{\sqrt{A^2+B^2}},\frac{  B}{\sqrt{A^2+B^2}} \left )
    With this vector the equation of the line gives zero when P(x,y) on the line
    and the signed distance of P from the line if it is not on the line.
    So this distance
    \displaystyle \left |\frac{Ax+By+C}{\sqrt{A^2+B^2}} \right |=d
    Wow, thanks ever so much! I also have another question, is it okay if I PM you instead of making another thread?

    Also what does the 'constant distance of five units' mean in the question?

    Thanks again!
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    (Original post by YdLeet)
    Wow, thanks ever so much! I also have another question, is it okay if I PM you instead of making another thread?

    Thanks again!
    OK
 
 
 
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