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    The vertices of a triangle are A(2,4), B(-8,2) and C(4,-6). The points P and Q are the midpoint of AB and AC respectively. The line through P perpendicular to AB meets the line through Q perpendicular to AC at R.
    1. Show that the equation of PR is y+5x+12=0 and find the equation of QR.
    2. Find the co-ordinates of R.
    3. Show that BR=CR

    So these are the working I've got, and I'm a little confused.
    Midpoint AB =  (\frac{2-8}{2} , \frac{4+2}{2})
    Midpoint AB = P
    P = (-3,3)
    Midpoint AC =  (\frac{2+4}{2} , \frac{4-6}{2})
    Midpoint AC = Q
    Q = (3,-1)
    Gradient AB =  (\frac{2-4}{-8-2})
    Gradient AB =  \frac{1}{5}
    Perpendicular to AB =  3=-5(-3)+ c
    So this means PR =  y=-5x-12 

 0=y+5x+12 So is this okay for a?
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    (Original post by Abusive)
    The vertices of a triangle are A(2,4), B(-8,2) and C(4,-6). The points P and Q are the midpoint of AB and AC respectively. The line through P perpendicular to AB meets the line through Q perpendicular to AC at R.
    1. Show that the equation of PR is y+5x+12=0 and find the equation of QR.
    2. Find the co-ordinates of R.
    3. Show that BR=CR


    So these are the working I've got, and I'm a little confused.
    Midpoint AB =  (\frac{2-8}{2} , \frac{4+2}{2})
    Midpoint AB = P
    P = (-3,3)
    Midpoint AC =  (\frac{2+4}{2} , \frac{4-6}{2})
    Midpoint AC = Q
    Q = (3,1)
    Gradient AB =  (\frac{-8-2}{2-4})
    Gradient AB = 5
    Perpendicular to AB =  3=\frac{-1}{5} (\frac{-3}{1}) + c
    So this means PR =  3+5(\frac{-3}{1})+12
    Now how do I answer the questions?
    You have your coordinates of P correct. You almost got Q - it should be Q(3,-1) which is what I think you meant.

    Next you need the gradient AB which you've worked out incorrectly - it should be \frac{2-4}{-8-2} - and you want the perpendicular gradient to AB after you find this. Construct the equation through P, perpendicular to AB, as shown in the question. This is your line \ell_1.

    Next do the same thing at Q. You know it's coordinate, now you need the gradient of AC and get the perpendicular gradient to it before constructing \ell_2 which is perpendicular to AC through Q.

    R is where \ell_1 and \ell_2 intersect, I assume you know how to find R??

    The for part C it's just applying Pythagoras' Theorem to show distances are equal.
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    (Original post by RDKGames)
    You have your coordinates of P correct. You almost got Q - it should be Q(3,-1) which is what I think you meant.

    Next you need the gradient AB which you've worked out incorrectly - it should be \frac{2-4}{-8-2} - and you want the perpendicular gradient to AB after you find this. Construct the equation through P, perpendicular to AB, as shown in the question. This is your line \ell_1.

    Next do the same thing at Q. You know it's coordinate, now you need the gradient of AC and get the perpendicular gradient to it before constructing \ell_2 which is perpendicular to AC through Q.

    R is where \ell_1 and \ell_2 intersect, I assume you know how to find R??

    The for part C it's just applying Pythagoras' Theorem to show distances are equal.
    Hey, I've updated the original post. Would that suffice as an answer for a?
    Also yes I know how to do the rest I'm sure. Thanks for your help!
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    (Original post by Abusive)
    Hey, I've updated the original post. Would that suffice as an answer for a?
    Also yes I know how to do the rest I'm sure. Thanks for your help!
    It's hard to read as it stands but if you're essentially taking y=mx+c and substituting m=-5 with (-3,3) before solving for c, then you're fine. Except I'm not sure why you're saying "3=-5(-3)-12 so 0=15-12-3" as that is not needed and doesn't show anything.

    Once you derive y+5x+12=0 that's all there is to it.
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    (Original post by RDKGames)
    It's hard to read as it stands but if you're essentially taking y=mx+c and substituting m=-5 with (-3,3) before solving for c, then you're fine. Except I'm not sure why you're saying "3=-5(-3)-12 so 0=15-12-3" as that is not needed and doesn't show anything.

    Once you derive y+5x+12=0 that's all there is to it.
    Thanks for your help means a lot.
 
 
 
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