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# C1 Coordinate Geometry, what is the reason for doing this step in the solution? watch

1. Solution is here:

I completely understand the first point. When you work out where something crosses the x axis, you make y=0, and vice versa - to work out where it crosses the y axis you make x=0, and then solve the equation. I then understand that you have the coordinate (3,0), which is where the line crosses the x axis.

I attempted the question myself without looking at the solution and got the same method and steps, but I am trying to make a mental note and explain why you use the formula y-y1=m(x-x1) in this instance after you have the equation in the form y=mx+c. What is the reason as to why after you have it as y=mx+c, you then sub the gradient and coordinates into that equation?

Thanks!
2. (Original post by blobbybill)
Solution is here:

I completely understand the first point. When you work out where something crosses the x axis, you make y=0, and vice versa - to work out where it crosses the y axis you make x=0, and then solve the equation. I then understand that you have the coordinate (3,0), which is where the line crosses the x axis.

I attempted the question myself without looking at the solution and got the same method and steps, but I am trying to make a mental note and explain why you use the formula y-y1=m(x-x1) in this instance after you have the equation in the form y=mx+c. What is the reason as to why after you have it as y=mx+c, you then sub the gradient and coordinates into that equation?

Thanks!
y-y1=m(x-x1) is the equation of a straight line if you know the gradient m and a point that the line passes through (x1, y1).

This question wants you to find the equation of a straight line given gradient 2/3 which passes through A so it would make sense to use y-y1=m(x-x1).

I've got a feeling though that this won't address your confusion. If you're still unsure, can you please post how you would have done this question so we can see your thought process.
3. (Original post by blobbybill)
Solution is here:

I completely understand the first point. When you work out where something crosses the x axis, you make y=0, and vice versa - to work out where it crosses the y axis you make x=0, and then solve the equation. I then understand that you have the coordinate (3,0), which is where the line crosses the x axis.

I attempted the question myself without looking at the solution and got the same method and steps, but I am trying to make a mental note and explain why you use the formula y-y1=m(x-x1) in this instance after you have the equation in the form y=mx+c. What is the reason as to why after you have it as y=mx+c, you then sub the gradient and coordinates into that equation?

Thanks!
You cannot simply sub it into because you do not know the intersection on the y-axis for this line - this would be your which is unknown otherwise.

By using you are translating a line of gradient onto the point in order to go through it. Furthermore, rearrangement of this equation shows that you get and keep in mind that represents two constants which, when added, make up an entirely different constant that call hence where it comes from. So you can't really use the form right off the bat with the gradient and a point because you do not know where this wanted line crosses the y-axis.
4. (Original post by blobbybill)
I attempted the question myself without looking at the solution and got the same method and steps, but I am trying to make a mental note and explain why you use the formula y-y1=m(x-x1) in this instance after you have the equation in the form y=mx+c. What is the reason as to why after you have it as y=mx+c, you then sub the gradient and coordinates into that equation?

Thanks!
Well you don't have the equation of the line in the form y=mx+c.

You have the equation of the original line in that form. And that's gotten you to the stage where you've found the point, A.

Now you're looking for a new line with a different gradient, and you have a point, A. Hence the method
5. (Original post by ghostwalker)
Well you don't have the equation of the line in the form y=mx+c.

You have the equation of the original line in that form. And that's gotten you to the stage where you've found the point, A.

Now you're looking for a new line with a different gradient, and you have a point, A. Hence the method
(Original post by RDKGames)
You cannot simply sub it into because you do not know the intersection on the y-axis for this line - this would be your which is unknown otherwise.

By using you are translating a line of gradient onto the point in order to go through it. Furthermore, rearrangement of this equation shows that you get and keep in mind that represents two constants which, when added, make up an entirely different constant that call hence where it comes from. So you can't really use the form right off the bat with the gradient and a point because you do not know where this wanted line crosses the y-axis.
(Original post by notnek)
y-y1=m(x-x1) is the equation of a straight line if you know the gradient m and a point that the line passes through (x1, y1).

This question wants you to find the equation of a straight line given gradient 2/3 which passes through A so it would make sense to use y-y1=m(x-x1).

I've got a feeling though that this won't address your confusion. If you're still unsure, can you please post how you would have done this question so we can see your thought process.
It is possible to use y=mx+c by writing y=2/3x+c then subbing x=3, y=0 to give 0 = 2+c -> c = -2, so y = 2/3x-2.
In fact, this method is usually taught first in A-Level courses, because y-y1=m(x-x1) is a new result, whereas y=mx+c is already known from GCSE.
6. (Original post by HapaxOromenon3)
It is possible to use y=mx+c by writing y=2/3x+c then subbing x=3, y=0 to give 0 = 2+c -> c = -2, so y = 2/3x-2.
In fact, this method is usually taught first in A-Level courses, because y-y1=m(x-x1) is a new result, whereas y=mx+c is already known from GCSE.
was waiting for this
7. (Original post by HapaxOromenon3)
It is possible to use y=mx+c by writing y=2/3x+c then subbing x=3, y=0 to give 0 = 2+c -> c = -2, so y = 2/3x-2.
In fact, this method is usually taught first in A-Level courses, because y-y1=m(x-x1) is a new result, whereas y=mx+c is already known from GCSE.
No need to quote all of us - we're all aware of this!

To me it seemed like the OPs question wasn't to do with using y-y1... instead of the y=mx+c method, but some other misunderstanding. I wasn't sure which is why I asked for clarification.

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