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    Find the shortest distance between the point (9,-3) and the line y=x

    okay I get this be be root 90 by doing

    l = sqrt (0-9)^2 + (0-(-3))^2

    the answer is 2 root 6 according to my book

    sure the coordinates of y=x is (0,0)?

    thanks
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    Useful Fact #2: The shortest distance between a point and a line is the perpendicular from the point to the line.

    The base of the perpendicular is not (0,0).
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    (Original post by madfish)
    Find the shortest distance between the point (9,-3) and the line y=x

    okay I get this be be root 90 by doing

    l = sqrt (0-9)^2 + (0-(-3))^2

    the answer is 2 root 6 according to my book

    sure the coordinates of y=x is (0,0)?

    thanks
    y = x is not a point, it is a line.
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    (Original post by aznkid66)
    Useful Fact #2: The shortest distance between a point and a line is the base of the perpendicular from the point to the line.

    The base of the perpendicular is not (0,0).
    how do we find the base of the perpidicular?
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    Note that the perpendicular extended is a new line that is perpendicular to the first. In addition, this new line crosses through the aforementioned point. The base of the perpendicular is conveniently the one point where these two lines meet.
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    For reference:

    Attachment 204703
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    (Original post by aznkid66)
    For reference:

    Attachment 204703
    your help is really appreciated

    but it wont let me view you image

    how do we find the perpendicular base coordinates?
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    To rephrase my previous hint, you want to find the intersection between L1: y=x and L2 (the line of the perpendicular bisector). Because of this, you need to first find the equation of L2.

    Useful Fact #4: You can find the equation of any line given its slope and a point on the line.

    To find L2's slope, note that L2 is perpendicular to L1: y=x.
    As for a point, note that the perpendicular extends from the point (9,-3).
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    (Original post by aznkid66)
    To rephrase my previous hint, you want to find the intersection between L1: y=x and L2 (the line of the perpendicular bisector). Because of this, you need to first find the equation of L2.

    Useful Fact #4: You can find the equation of any line given its slope and a point on the line.

    To find L2's slope, note that L2 is perpendicular to L1: y=x.
    As for a point, note that the perpendicular extends from the point (9,-3).
    but how do we know they are perpendicular or even touching? It doesn't tell us in the question so why do we make this assumption?
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    It's a commonly known fact. For empirical evidence, draw a line and a point and the perpendicular from the point to the line. Can you draw a line that is shorter than the one you just drew?
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    There is a certain formula for calculating the distance between a point and line. That would be: |aX + bY + c|/sqrt(a^+b^).

    Where the line is ax+by+c=0 (so thats where you get the a,b & c from) and X&Y are the coordinates of your point A(x,y)

    Hope that helps

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    (Original post by madfish)
    Find the shortest distance between the point (9,-3) and the line y=x

    okay I get this be be root 90 by doing

    l = sqrt (0-9)^2 + (0-(-3))^2

    the answer is 2 root 6 according to my book

    sure the coordinates of y=x is (0,0)?

    thanks
    The line y=x ISN'T a single point (0,0) - it's the set of points (x,x)!

    You can adapt your method by working out

    l^2 = (x-9)^2 + (x-(-3))^2

    and then minimizing the quadratic you get on the RHS

    (BTW: are you sure the answer is 2root6 ?)
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    Right, minimizing the quadratic is probably the better method for this question. Just checking, do you know how to find the y value of a quadratic's vertex?
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    (Original post by davros)
    The line y=x ISN'T a single point (0,0) - it's the set of points (x,x)!

    You can adapt your method by working out

    l^2 = (x-9)^2 + (x-(-3))^2

    and then minimizing the quadratic you get on the RHS

    (BTW: are you sure the answer is 2root6 ?)
    I believe it's 6 root 2, he probably got it backwards.
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    (Original post by justinawe)
    I believe it's ...
    That's what I got too - I didn't want to offer up the answer until the OP had had another go
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    (Original post by davros)
    That's what I got too - I didn't want to offer up the answer until the OP had had another go
    Well, it was the answer the back of his book, not his answer, so I thought it would be appropriate to offer up the correct answer. It was probably just a typo on his part, unless the writer of the book made an error.
 
 
 
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