HELP! Equations of lines

Ok I hate graph and line questions and I'm really confused on this question.

1. Find the equations of these lines. Give your answers in the form ax+by+c=0 where a,b,c are integers

a) A line with the gradient -4 that passes through the point A(5,0)
b) A line with the gradient 1/2 that passes through the point A(2,3)

Help please!

Your gradient is -a/b so you can put those in straight away

Then put your point in for c

Here is an example
Gradient 3/4 passes through (1,3)

-a/b = 3/4 so a=-3 and b=4 so -3x + 4y + c = 0
(x,y) = (1,3) so -3+12+c=0 so c=9

I'm a little confused where you say gradient is -a/b
For the first one, would it be -4x+16y+c=0 when you sub in the -a/b bit?
My brain is mush atm and none of this is making sense

I'm a little confused where you say gradient is -a/b
For the first one, would it be -4x+16y+c=0 when you sub in the -a/b bit?
My brain is mush atm and none of this is making sense

Where has the 16 come from

-4 is -4/1


Are you happier with y=mx+c

Or

Y-y1 = m(x-X1)

If you are ... Use one of those and then re arrange

Oh.. my bad. The book says that the answer is 4x+y-20=0 but wouldn't the 4 be negative?

I'm a little confused where you say gradient is -a/b
For the first one, would it be -4x+16y+c=0 when you sub in the -a/b bit?
My brain is mush atm and none of this is making sense

No

-4/1 = -a/b

So a is 4 and b is 1

Just do y=mx + c where m is the gradient given, then plug in y and x to find c less to remember and less faffing with a's and b's! Then rearrange for the requested form.


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I understand it now! Thanks for your help. It's much easier now that I know about y-b=m(x-a) and rearranging it

I've never understood that method to be honest! Even in year13 i still never use any of the C1 or GCSE coordinate geometry methods with all the different confusing formulae! But if they work for you, stick with them

Just pick what works best for you! There's different methods and they all result in the same answer in the end, it's just about finding what works best for you.


Posted from TSR Mobile

If you have coordinates $(x_1,y_1)$ and $(x_2,y_2)$, you can work out the gradient, m, by using the formula $m=\dfrac{y_2-y_1}{x_2-x_1}$. If we substitute $(x,y)$ for $(x_2,y_2)$ the equation for the gradient becomes $m=\dfrac{y-y_1}{x-x_1}$. Rearranging gives the required result: $y-y_1=m(x-x_1)$.

I know the method, I just avoid it like the plague. My one teacher was surprised that my class avoided those methods, i suppose it's exactly the same, we just prefer not using that formula and do it the method i mentioned, which, as I said is the same essentially.


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I know the method, I just avoid it like the plague. My one teacher was surprised that my class avoided those methods, i suppose it's exactly the same, we just prefer not using that formula and do it the method i mentioned, which, as I said is the same essentially.


Posted from TSR Mobile

I'd always go with the y-y1 = m (x-x1) where y1 and x1 are ANY point on the line. Its far easier and less likely to make a mistake

If you say so :P
I'd much rather stick with what I know though

Ok I hate graph and line questions and I'm really confused on this question.

1. Find the equations of these lines. Give your answers in the form ax+by+c=0 where a,b,c are integers

a) A line with the gradient -4 that passes through the point A(5,0)
b) A line with the gradient 1/2 that passes through the point A(2,3)

Help please!