Equations Of Straight Line
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.
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.
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.
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.
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.
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.
Original post by TeeEm
am I too late?
**** off.
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.
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?
Original post by constellarknight
**** off.
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.
Original post by TeeEm
am I too late?
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