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Math Help - Perpendicular Distance Question

  1. #1
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    Smile Perpendicular Distance Question

    How would I do this?

    Find the equation of the lines that pass through the point (-3, -4) and that are a distance of √10 from the point (2,1).

    I believe I need to use this formula to solve the question:
    |ax1 + by1 + c| / √(a + b)

    Probably isn't hard but any help would be greatly appreciated!
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Perpendicular Distance Question

    I would approach this problem by finding the lines through (-3,-4) and tangent to the circle:

    (x-2)^2+(y-1)^2=10

    Using the point-slope formula, let the tangent lines have the form:

    y+4=m(x+3)

    y=m(x+3)-4

    Now, substitute into the circle for y to get a quadratic in x, and set the discriminant to zero (do you know why?) to get a quadratic in m.
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  3. #3
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    Re: Perpendicular Distance Question

    Thanks a million for the help but I'm totally confused by that sorry, I don't really get any of that! I'm doing some Co-ordinate Geometry homework and the topic is 'The Perpendicular Distance from a point to a line' and we haven't come across anything to do with tangents or circles so I think we're expected to do it that way. Is there any other way you can think of to do it?
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    Re: Perpendicular Distance Question

    firstly find the eqation of a line passing thru (-3,-4) by point slope form then apply formula of a dist b/w pont and a line to find m after that put the value of m to get the required equation
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  5. #5
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    Re: Perpendicular Distance Question

    point slope form
    y-y1=m(x-x1)
    y+4=m(x+3)
    i.e. mx-y+3m-4=0--------- I
    then apply perpendicular dist b/w point and a line
    |ax1 + by1 + c| / √(a + b) by equating tht with √10
    find m from here then put tht in I
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Perpendicular Distance Question

    Quote Originally Posted by HelenMc9 View Post
    Thanks a million for the help but I'm totally confused by that sorry, I don't really get any of that! I'm doing some Co-ordinate Geometry homework and the topic is 'The Perpendicular Distance from a point to a line' and we haven't come across anything to do with tangents or circles so I think we're expected to do it that way. Is there any other way you can think of to do it?
    Sorry, I was just musing about how I might approach this problem from a different angle.

    As I showed in my first post, the tangent lines will have the form:

    y+4=m(x+3)

    So, let's put that into standard form:

    -mx+y+(4-3m)=0

    So, for the standard form of a line we identify:

    A=-m,\,B=1,\,C=4-3m

    Now, our formula for the distance d between a point (x_1,y_1) and the line Ax+By+C=0 is:

    d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+C^2}}

    We are given:

    d=\sqrt{10},\,(x_1,y_1)=(2,1) and so we have:

    \sqrt{10}=\frac{|-2m+1+4-3m|}{\sqrt{(-m)^2+(1)^2}}

    \sqrt{10}=\frac{5|m-1|}{\sqrt{m^2+1}}

    Now, square the equation:

    10=\frac{25(m-1)^2}{m^2+1}

    Now, solve for m. You will get two values, corresponding to the two possible lines.

    What do you find?
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  7. #7
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    Re: Perpendicular Distance Question

    10m + 10 = 25(m-2m+1)
    10m + 10 = 25m -50m + 25
    15m-50m+15=0
    3m-10m+3=0
    3m-9m-m+3=0
    3m(m-3)-1(m-3) = 0
    (3m-1)(m-3) = 0
    3m=1 or m=3

    so m=1/3 or m=3. Is that right?


    And then I suppose I can fill in the two separate values for m into -mx + y + (4-3m) = 0 to get the two separate equations?
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Perpendicular Distance Question

    Yes, good work!
    Thanks from HelenMc9
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  9. #9
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    Re: Perpendicular Distance Question

    Thank-you very much! I understand it now
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: Perpendicular Distance Question

    Glad to help, particularly for someone who is a math helper's dream...you stated the problem clearly in its entirety with your thoughts on what you needed to use, then responded to feedback with work too. That is rare.
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