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    The straight lines 5x+4y+2=0 and 15x+ay+8=0 are parallel. Find a.

    The straight lines 6x-4y+3=0 and 3x-ky+8=0 are parallel. Find k.

    Can someone help with how to do these? Very urgent, thanks.
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    (Original post by ross2000)
    The straight lines 5x+4y+2=0 and 15x+ay+8=0 are parallel. Find a.

    The straight lines 6x-4y+3=0 and 3x-ky+8=0 are parallel. Find k.

    Can someone help with how to do these? Very urgent, thanks.
    Get them into the form y = mx + c, then use the fact that parallel lines have the same gradient.

    Post all your thoughts/working if you get stuck.
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    (Original post by ross2000)
    The straight lines 5x+4y+2=0 and 15x+ay+8=0 are parallel. Find a.

    The straight lines 6x-4y+3=0 and 3x-ky+8=0 are parallel. Find k.

    Can someone help with how to do these? Very urgent, thanks.
    5x+4y+2=0 -> 4y=-5x-2 -> y=-(5/4)x-1/2, so the gradient of this line has gradient -5/4.
    15x+ay+8=0 -> ay=-15x-8 -? y=-(15/a)x-8/a, so the gradient of this line is -15/a.
    Thus -5/4=-15/a -> 15/a=5/4=15/12, so a=12. The method for the second question is the same.
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    am I too late?
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    (Original post by TeeEm)
    am I too late?
    **** off.
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    (Original post by notnek)
    Get them into the form y = mx + c, then use the fact that parallel lines have the same gradient.

    Post all your thoughts/working if you get stuck.
    Should it be y-b=m(x-a) as there is no y intercept given? Also when subtracting both sides by y would it be ay or just y?
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    (Original post by constellarknight)
    **** off.
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    (Original post by ross2000)
    Should it be y-b=m(x-a) as there is no y intercept given? Also when subtracting both sides by y would it be ay or just y?
    See my post above. The point is you can algebraically rearrange it into the y=mx+c form, which allows you to easily see the gradients.
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    (Original post by TeeEm)
    am I too late?
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    I got it, thank you all
 
 
 
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