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    question b) is the one I do not know how to approach. I know I could probably use pythagoras, but I predict that the calculations will get a little bit messy. There are other ways to complete this on the mark scheme, like "gradient LM * gradient MN", but I don't understand where they're coming from.

    any help pls?
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    No Pythagoras involved. The line MN passes through M and is perpendicular to ML. Find the equation of the line passing through both M and N and then you should see how to find p.

    To find p, plug x=16 I to the equation of the line MN that you found.
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    if you know the gradient of LM you can find the gradient of MN as it is perpendicular...
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    (Original post by frostyy)


    question b) is the one I do not know how to approach. I know I could probably use pythagoras, but I predict that the calculations will get a little bit messy. There are other ways to complete this on the mark scheme, like "gradient LM * gradient MN", but I don't understand where they're coming from.

    any help pls?
    If (and only if) two lines are perpendicular, the product of their gradients is -1. (You can prove this using vectors).
    Thus gradient LM = (-4-2)/(7+1) = -6/8 = -3/4, and gradient MN = (p+4)/(16-7) = (p+4)/9.
    Since angle LMN = 90 degrees, LM is perpendicular to MN.
    Thus -3/4 * (p+4)/9 = -1 -> 3(p+4)/36 = 1 -> p+4 = 12 -> p=8.
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    (Original post by HapaxOromenon2)
    If (and only if) two lines are perpendicular, the product of their gradients is -1. (You can prove this using vectors).
    Thus gradient LM = (-4-2)/(7+1) = -6/8 = -3/4, and gradient MN = (p+4)/(16-7) = (p+4)/9.
    Since angle LMN = 90 degrees, LM is perpendicular to MN.
    Thus -3/4 * (p+4)/9 = -1 -> 3(p+4)/36 = 1 -> p+4 = 12 -> p=8.
    Okay, thank you, I get that.

    What about part c)? I know the answer is 2 + p + 4, but I'm not sure why?
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    (Original post by frostyy)
    Okay, thank you, I get that.

    What about part c)? I know the answer is 2 + p + 4, but I'm not sure why?
    Does this help at all:

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    (Original post by Zacken)
    Does this help at all:

    That's how I illustrated it to myself at the start, but I still can't see why you'd add the present 3 y-coords (+ ignore the minus sign at M (7, -4)?).
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    (Original post by frostyy)
    That's how I illustrated it to myself at the start, but I still can't see why you'd add the present 3 y-coords (+ ignore the minus sign at M (7, -4)?).
    Would it help if you think about it as vectors? (start from M, add the vector MN then add the vector -ML)

    NB: Another way of doing this would be to write K(x,y) then use the gradients perpendicular to LM and parallel to MN to get a simultaneous equation in x,y.
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    (Original post by Zacken)
    Would it help if you think about it as vectors? (start from M, add the vector MN then add the vector -ML)
    Okay, that makes sense, thank you. Woah, I'd never expect to use vectors in C1.
 
 
 
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