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# Equations Of Straight Line watch

1. 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.
2. (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.
3. (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.
4. am I too late?
5. (Original post by TeeEm)
am I too late?
**** off.
6. (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?
7. (Original post by constellarknight)
**** off.
8. (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.
9. (Original post by TeeEm)
am I too late?
10. I got it, thank you all

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