methods 1 2 resources Maths Methods Questions

[brightsky]

The post above was directed at your question above. You'll get two types of questions:

1. Given some information about the line, find the equation of line.
2. Given the equation of the line, sketch the line.

The 'rise over run' method, if you wish to call it that, is only really used for Type 1 questions. The question you are posing now is a Type 2 question.

Before you appeal to any random formulas, think about what you can do intuitively. You have the equation of a line. This literally tells you EVERYTHING about the line. Everything!

Suppose you were asked to sketch y = 1/2*x^2*e^(sqrt(pi*x) instead. No one knows what this graph looks like. No one knows any formulas which can be used to work out the shape of the graph instantaneously. But the equation tells you EVERYTHING. All you need to do is laboriously plug in x values, find corresponding y-values and then plot the point on a Cartesian graph. After you've repeated this procedure a billion times, you'll have a billion points. If you join these points up, you'll have the graph of the function.

Now, your question is considerably easier than the question I posed. You know, from your studies, that y= 2x - 8 is a line. Do you really need to plot a billion points to determine the shape of the graph? No. As discussed above, two points is enough to uniquely define a straight line. So, all you need to do is find two points on the graph and then join those points up. I think this is what you were getting at when you mentioned the 'two point' method.

To show you. We have y = 2x - 8. Substitute x = 50. We have y = 92. We know that (50,92) is on the graph. We have one point. Okay, now we need another. Substitute x = 31. We have y = 54. We know that (31,54) is on the graph. Plot these points on grid paper and then join them. You'll get the graph you want.

It is usually easier to just substitute x = 0, and x = 1 or something small like that. But what's stopping you from substituting x = 12381230981203984? Your rationality probably. But the principle remains.

[methods 1 2 resources]

E.g y=2x+6 which method do i use and you guys say it doesn't matter but if i use two different methods i ill get two different answers?

[Stevensmay]

Then you need to find some more resources and do practical questions so that you do understand.

And in a SAC you will be rarely told how to do something. The answer will not be different depending on what method you use, I advise you to check your working.

[brightsky]

I'd also like to add that set methods/techniques are overrated. Consider the following problem:

Prove that every even integer greater than 2 can be expressed as the sum of two primes.

What 'method' do we use to solve this problem? Well, the method hasn't been invented yet. There is no method. You just need to use your intuition and a little independent thinking. (Perhaps the adjective 'little' is unwarranted since this problem hasn't actually been solved yet.)

In short, brain over method.

[methods 1 2 resources]

The post above was directed at your question above. You'll get two types of questions:

1. Given some information about the line, find the equation of line.
2. Given the equation of the line, sketch the line.

The 'rise over run' method, if you wish to call it that, is only really used for Type 1 questions. The question you are posing now is a Type 2 question.

Before you appeal to any random formulas, think about what you can do intuitively. You have the equation of a line. This literally tells you EVERYTHING about the line. Everything!

Suppose you were asked to sketch y = 1/2*x^2*e^(sqrt(pi*x) instead. No one knows what this graph looks like. No one knows any formulas which can be used to work out the shape of the graph instantaneously. But the equation tells you EVERYTHING. All you need to do is laboriously plug in x values, find corresponding y-values and then plot the point on a Cartesian graph. After you've repeated this procedure a billion times, you'll have a billion points. If you join these points up, you'll have the graph of the function.

Now, your question is considerably easier than the question I posed. You know, from your studies, that y= 2x - 8 is a line. Do you really need to plot a billion points to determine the shape of the graph? No. As discussed above, two points is enough to uniquely define a straight line. So, all you need to do is find two points on the graph and then join those points up. I think this is what you were getting at when you mentioned the 'two point' method.

To show you. We have y = 2x - 8. Substitute x = 50. We have y = 92. We know that (50,92) is on the graph. We have one point. Okay, now we need another. Substitute x = 31. We have y = 54. We know that (31,54) is on the graph. Plot these points on grid paper and then join them. You'll get the graph you want.

It is usually easier to just substitute x = 0, and x = 1 or something small like that. But what's stopping you from substituting x = 12381230981203984? Your rationality probably. But the principle remains.

i'm so confused.  You guys are telling me it doesn't matter what method i use then you guys says it does matter as you can determine what method you use from the gradient.  if a question was given y=2x+8, there are 5 methods to choose from how do i know which method to choose from?  And stevensmay said they all give the same answer know they don't?

[Stevensmay]

And stevensmay said they all give the same answer know they don't?

No matter what method I use I will always get the line shown below. There is absolutely no other way to graphically express the equation y = 2x+ 8, regardless of what method I use to sketch it.

Spoiler

Spoiler

Lets say I wanted to build a house. This house will always end up the same no matter how I build it.
If I pick the easy method of house building, I put the floor in, then the walls then the roof. That was easy.

If I pick the hard method to build the house, I might hold the roof up with cranes, then put the floor in and try to squeeze the walls in. Even though I end up with the same house, this was a lot harder to do.

[lzxnl]

i'm so confused.  You guys are telling me it doesn't matter what method i use then you guys says it does matter as you can determine what method you use from the gradient.  if a question was given y=2x+8, there are 5 methods to choose from how do i know which method to choose from?  And stevensmay said they all give the same answer know they don't?

All methods lead to the same answer. Some methods may give a result that looks different; that is true, but you should always be able to rearrange them to look identical.

Let's break down the thought processes that go through my mind when I see a straight line.

1. Oh look, it looks like a straight line. I know what these look like.
2. Hmm. I only need two points to define this straight line (pretty much a definition)
3. Intercepts are needed anyway, so why don't I just find those?
4. Oh wait...there are x and y intercepts; that makes two points.
5. Found intercepts, drew line through them; that's my line.

A SAC really shouldn't ask how to draw a straight line. They may ask you to use a table of values; that is possible, but you'll be able to see from a table given.

If you're given two points, however, and you want to find the equation of the straight line, you have a different job.
As brightsky has mentioned, you find the gradient first, then sub into y-y1 = m(x-x1). I'll explain why you need to do this.
The reason is fairly simple. A straight line can be expressed in the form y=mx+c. If you can find m, then c will be easy to find.
So, you look at the points. Gradient is defined as rise/run (something you unfortunately have to remember). What's the rise? Well the rise is the change in the y coordinate. The run is the change in the x coordinate. Say you have (1,2) and (0,6). The change in the y coordinate as you go from (1,2) to (0,6) is +4 as it increases. The change in the x coordinate as you go from (1,2) to (0,6) is -1, as the x coordinate decreases. Dividing, you get rise/run = +4/-1 = -4. That's your gradient.

Now, where does this y-y1 = m(x-x1) formula come from? Well, if this is the formula for the straight line, then it must satisfy two requirements:
1. The gradient is correct. Here, you can rearrange to get y = m(x-x1) + y1, and you can see that the gradient is m, which is what you wanted. That requirement is met.
2. The line actually passes through the given point. Well, if x=x1 and y=y1, the left hand side becomes y1-y1=0 and the right hand side is m*(x1-x1) = 0. As both sides are equal, the given point satisfies the equation of the straight line. Thus, the straight line formula works.

OK, so we've been through that trouble to find the gradient. y-y1 = m(x-x1). For my example, m = -1. We'll use (0,6)
x1=0 y1=6
y-8 = -x
y = -x + 8
And there we have it; the equation of the line.

[methods 1 2 resources]

All methods lead to the same answer. Some methods may give a result that looks different; that is true, but you should always be able to rearrange them to look identical.

Let's break down the thought processes that go through my mind when I see a straight line.

1. Oh look, it looks like a straight line. I know what these look like.
2. Hmm. I only need two points to define this straight line (pretty much a definition)
3. Intercepts are needed anyway, so why don't I just find those?
4. Oh wait...there are x and y intercepts; that makes two points.
5. Found intercepts, drew line through them; that's my line.

A SAC really shouldn't ask how to draw a straight line. They may ask you to use a table of values; that is possible, but you'll be able to see from a table given.

If you're given two points, however, and you want to find the equation of the straight line, you have a different job.
As brightsky has mentioned, you find the gradient first, then sub into y-y1 = m(x-x1). I'll explain why you need to do this.
The reason is fairly simple. A straight line can be expressed in the form y=mx+c. If you can find m, then c will be easy to find.
So, you look at the points. Gradient is defined as rise/run (something you unfortunately have to remember). What's the rise? Well the rise is the change in the y coordinate. The run is the change in the x coordinate. Say you have (1,2) and (0,6). The change in the y coordinate as you go from (1,2) to (0,6) is +4 as it increases. The change in the x coordinate as you go from (1,2) to (0,6) is -1, as the x coordinate decreases. Dividing, you get rise/run = +4/-1 = -4. That's your gradient.

Now, where does this y-y1 = m(x-x1) formula come from? Well, if this is the formula for the straight line, then it must satisfy two requirements:
1. The gradient is correct. Here, you can rearrange to get y = m(x-x1) + y1, and you can see that the gradient is m, which is what you wanted. That requirement is met.
2. The line actually passes through the given point. Well, if x=x1 and y=y1, the left hand side becomes y1-y1=0 and the right hand side is m*(x1-x1) = 0. As both sides are equal, the given point satisfies the equation of the straight line. Thus, the straight line formula works.

OK, so we've been through that trouble to find the gradient. y-y1 = m(x-x1). For my example, m = -1. We'll use (0,6)
x1=0 y1=6
y-8 = -x
y = -x + 8
And there we have it; the equation of the line.

I'm still confused, can you give me a question without telling me what method to use and ill answer it

[Stevensmay]

Find the gradient and intercepts of y = 4x-2.


Find the equation to the line that passes through (1,1) and (4,4)

Find the equation to the line that has gradient 6 and y-intercept of -4.

[methods 1 2 resources]

Find the gradient and intercepts of y = 4x-2.


Find the equation to the line that passes through (1,1) and (4,4)

Find the equation to the line that has gradient 6 and y-intercept of -4.

1.)  M=4x intercepts=x= 2 y=-2

2.) 1x+0 or just leave as 1x i guess

3.)6x-4


Am i right?

[lzxnl]

1.)  M=4x intercepts=x= 2 y=-2

2.) 1x+0 or just leave as 1x i guess

3.)6x-4


Am i right?

1. You just say m=4; no need for the x. Intercepts are correct

2. You need to say y=x, even though you know what you're doing.

3. OK, so you know how to do these questions. Just remember to put y= at the start, otherwise we don't know what the heck you're doing.


How about a more difficult one? A line passing through (1,1) and (4,4) is really easy to find as both coordinates are the same.
A straight line that passes through (1,2) and (5,7)?

[LOLs99]

1. You just say m=4; no need for the x. Intercepts are correct

2. You need to say y=x, even though you know what you're doing.

3. OK, so you know how to do these questions. Just remember to put y= at the start, otherwise we don't know what the heck you're doing.


How about a more difficult one? A line passing through (1,1) and (4,4) is really easy to find as both coordinates are the same.
A straight line that passes through (1,2) and (5,7)?

Is the x-int 0.5 or did i just wake up and my brain is still waking up    ... thanks

[lzxnl]

Yeah my mind is asleep. You're right.
Too much Tetris these days; can't think clearly anymore

[Stevensmay]

In addition to the one nliu posted.

3y = 2x + 10

For a nice challenge.
Find the gradient of the line perpendicular to y=3x + 2
Don't worry too much if you can't do that.

[brightsky]

In addition to the one nliu posted.

3y = 2x + 10

For a nice challenge.
Find the gradient of the line perpendicular to y=3x + 2
Don't worry too much if you can't do that.

Or rather, prove that if two lines with gradients m1 and m2 respectively were perpendicular, then m1*m2 = -1.