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    looking at the picture
    can some please tell me where does the author get the (-3) x 4 from.


    thank you Name:  ok.jpg
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    (Original post by bigmansouf)
    looking at the picture
    can some please tell me where does the author get the (-3) x 4 from.


    thank you Name:  ok.jpg
Views: 52
Size:  158.3 KB
    So for two lines to be perpendicular to each other, their gradients must times to equal -1.

    So for two general lines ax + by + c = 0 and dx + ey + f = 0, the gradients are -a/b and -d/e respectively.

    Therefore (-a/b) * (-d/e) = -1, which implies ad = - be or ad + be = 0

    In this example a = k+1, b = -3, d = k-2, e = 4

    Hence, (k+1)*(k-2) + (-3) *(4) = 0

    Hope this makes it clear where all the terms come from
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    (Original post by bigmansouf)
    looking at the picture
    can some please tell me where does the author get the (-3) x 4 from.


    thank you Name:  ok.jpg
Views: 52
Size:  158.3 KB
    This is how I think they have got it
    \left( k+1\right) x - 3y + 2 = 0 has gradient \frac{k+1}{3}
    and
    \left( k-2\right) x + 4y - 1 = 0 has gradient -\frac{k-2}{4}

    For the lines to be perpendicular:
    -\frac{k-2}{4}\times\frac{k+1}{3} =-1
    Therefore,
    \left(k-2\right)\left(k+1\right)=12
    \left(k-2\right)\left(k+1\right)-12=0
    \left(k-2\right)\left(k+1\right)+\left(-3\right)\times 4=0
    Don't know if this helps.
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    (Original post by xylas)
    So for two lines to be perpendicular to each other, their gradients must times to equal -1.

    So for two general lines ax + by + c = 0 and dx + ey + f = 0, the gradients are -a/b and -d/e respectively.

    Therefore (-a/b) * (-d/e) = -1, which implies ad = - be or ad + be = 0

    In this example a = k+1, b = -3, d = k-2, e = 4

    Hence, (k+1)*(k-2) + (-3) *(4) = 0

    Hope this makes it clear where all the terms come from
    thank you
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    (Original post by Cryptokyo)
    This is how I think they have got it
    \left( k+1\right) x - 3y + 2 = 0 has gradient \frac{k+1}{3}
    and
    \left( k-2\right) x + 4y - 1 = 0 has gradient -\frac{k-2}{4}

    For the lines to be perpendicular:
    -\frac{k-2}{4}\times\frac{k+1}{3} =-1
    Therefore,
    \left(k-2\right)\left(k+1\right)=12
    \left(k-2\right)\left(k+1\right)-12=0
    \left(k-2\right)\left(k+1\right)+\left(-3\right)\times 4=0
    Don't know if this helps.
    thanks you
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    For part two the roots of k^{2}-k+14 are k=\frac{1 \pm \sqrt{1+4\times14}}{2}=\frac{1 \pm \sqrt{57}}{2}

    And,
    \frac{1 + \sqrt{57}}{2}+\frac{1- \sqrt{57}}{2}=1
 
 
 
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