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C1 - I have a lines intersection question and written solution, can you explain it? watch

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    Question is:
    The points A and B have coordinates (-3, 8) and (5, 4) respectively. The straight line L1 passes through A and B.

    a) Find an equation for L1 giving your answer in the form ax + by + c =0, where a b and c are integers.
    b) Another straight line L2 is perpendicular to L1 and passes through the origin. Find an equation for L2.

    c) The lines L1 and L2 intersect at the point P. Use algebra to find the coordinates of P.
    I have attached the written solution. It is part C i don't understand how to do.

    Why do you substitute L2 into L1 in order to find where they intersect? The way I was taught to find points of intersection (with cubic graphs, not a question like this one) was to make the equation equal to each other, take one away from the other and solve from there.

    But I just don't understand why you substitute the equation for L2 (y=2x) into L1 (x+2y-13=0). Can somebody please explain why you do that?Thanks!

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    (Original post by blobbybill)

    Why do you substitute L2 into L1 in order to find where they intersect?

    Because at the point of intersection the x and y coordinates are the same for both lines. For the line y=2x you can express the y-coordinate as a function of x, since you know both y's are going to be the same, you can replace the y's in the other equation with 2x and solve for one variable.

    You can do the same and rearrange for x and substitute that one in and then solve for y instead if you wish.

    At the end once you find one variable, you find the other by plugging your known variable through one of the equations. It doesn't matter which because, well, the coordinates are the same after all so you get the same answer. Although it is useful to plug it through both because it is a method of verification and indication you got the question right.

    The way I was taught to find points of intersection (with cubic graphs, not a question like this one) was to make the equation equal to each other, take one away from the other and solve from there.
    That's still substitution if you think about it because you are equating y=f(x) to y=g(x) so replace the y in the first one you'd get g(x)=f(x) and solve from there.
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    (Original post by RDKGames)
    That's still substitution if you think about it because you are equating y=f(x) to y=g(x) so replace the y in the first one you'd get g(x)=f(x) and solve from there.
    What do you mean by g(x)? What is g(x)?

    Because at the point of intersection the x and y coordinates are the same for both lines. For the line y=2x you can express the y-coordinate as a function of x, since you know both y's are going to be the same, you can replace the y's in the other equation with 2x and solve for one variable.
    Ok, so why would you substitute one into the other rather than making them equal each other, taking one away from the other and then working from there?

    The markscheme (attached) for this question is the same one I used for previous questions regarding cubic graphs and points of intersection. With those questions, I took one away from the other, factorised it, worked out the x values and then substituted each x value into one of the line equations in order to find ther corresponding y coordinate at that point of intersection.

    I still don't get why you would just substitute one equation into the other (in this instance), whereas with something like points of intersection on a cubic graph, you take one equation away from the other, factorise it, work out the x values when y=0, then plug each x value into an equation to find the corresponding y coordinate.

    I haven't been taught this way (the way this mark scheme uses), nor have I ever seen an intersection question like this one before, so I don't know how to do it. I have only seen ones with cubic graphs, which I do a different method (much more longer winded) to find out the corresponding y value.

    Thanks so much, means a lot.
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    (Original post by blobbybill)
    What do you mean by g(x)? What is g(x)?
    Any other function of x.

    I.e. Let f(x)=x^3 and g(x)=2x+1 so intersections will be given by the condition where f(x)=g(x)


    Ok, so why would you substitute one into the other rather than making them equal each other, taking one away from the other and then working from there?
    Because you can't equate them in their given forms, how would you do it in this case?? If you wish to make them equal to each other, you would need to rearrange both equations for either x or y before you can make them equal to each other. But even then, this is still called substitution as per my example with f(x) and g(x).

    So following from the way you'd been taught, you would need to rearrange x+2y-13=0 \Rightarrow y=-\frac{1}{2}x+\frac{13}{2} and then equate this to y=2x which as I said is still called substitution.
    For simplicity and neater working however, it is much better to just substitute for one variable into the equation. This is particularly useful when it comes to higher order polynomials, logarithms, trigonometry, etc...

    The markscheme (attached) for this question is the same one I used for previous questions regarding cubic graphs and points of intersection. With those questions, I took one away from the other, factorised it, worked out the x values and then substituted each x value into one of the line equations in order to find ther corresponding y coordinate at that point of intersection.
    Please show me a good example where your method works out and I will perfectly point out at which stage you make a substitution just to make it clear.


    I still don't get why you would just substitute one equation into the other (in this instance), whereas with something like points of intersection on a cubic graph, you take one equation away from the other, factorise it, work out the x values when y=0, then plug each x value into an equation to find the corresponding y coordinate.
    There are various methods for obtaining a point of intersections; ie simultaneous equations, or substitution. It just so happens that substitution is the most common form of finding these intersections and you're allowed to substitute because the variables (in both equations) are equivalent at the points of intersections; therefore you are allowed to replace them.


    I haven't been taught this way (the way this mark scheme uses), nor have I ever seen an intersection question like this one before, so I don't know how to do it. I have only seen ones with cubic graphs, which I do a different method (much more longer winded) to find out the corresponding y value.

    Thanks so much, means a lot.
    Well you better get used to this as you do it throughout the entire A-Level maths and you need to understand WHY you do this, as I have explained above.
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    what was the answers to part a and b. ?i got a) y= -1/2x +10.5 b) y= -2x+18 need someone to check if this is right or notgradient was -1/2 but for the perpendicular one its -1/(-1/2) or reciprocal of -1/2 which is -2.
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    (Original post by Rajeevan)
    what was the answers to part a and b. ?i got a) y= -1/2x +10.5 b) y= -2x+18 need someone to check if this is right or notgradient was -1/2 but for the perpendicular one its -1/(-1/2) or reciprocal of -1/2 which is -2.
    Look at the mark scheme, both equations are there. Also -1/(-1/2) \not= -2
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    (Original post by blobbybill)
    Question is:
    The points A and B have coordinates (-3, 8) and (5, 4) respectively. The straight line L1 passes through A and B.

    a) Find an equation for L1 giving your answer in the form ax + by + c =0, where a b and c are integers.
    b) Another straight line L2 is perpendicular to L1 and passes through the origin. Find an equation for L2.

    c) The lines L1 and L2 intersect at the point P. Use algebra to find the coordinates of P.
    I have attached the written solution. It is part C i don't understand how to do.

    Why do you substitute L2 into L1 in order to find where they intersect? The way I was taught to find points of intersection (with cubic graphs, not a question like this one) was to make the equation equal to each other, take one away from the other and solve from there.

    But I just don't understand why you substitute the equation for L2 (y=2x) into L1 (x+2y-13=0). Can somebody please explain why you do that?Thanks!

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    You can.

    The person above me is trying to explain how it's easier to substitute. I think.

    You can set both equations equal to 0 and solve.

    I think though it's easier to substitute because it's quicker. I'd rather use a method i know to be quicker and the result x rather than use a longer method and still get the same result.

    Same principle, the 2 lines will intersect somewhere. The x and y co-ordinates will be equal somewhere along the lines. So if you substitute to get rid of y to leave you an equation with just x then you will find the x co-ordinate where both lines intersect. Substitution is just another form of saying these 2 things are equal.
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    (Original post by RDKGames)
    Any other function of x.

    I.e. Let f(x)=x^3 and g(x)=2x+1 so intersections will be given by the condition where f(x)=g(x)



    Because you can't equate them in their given forms, how would you do it in this case?? If you wish to make them equal to each other, you would need to rearrange both equations for either x or y before you can make them equal to each other. But even then, this is still called substitution as per my example with f(x) and g(x).

    So following from the way you'd been taught, you would need to rearrange x+2y-13=0 \Rightarrow y=-\frac{1}{2}x+\frac{13}{2} and then equate this to y=2x which as I said is still called substitution.
    For simplicity and neater working however, it is much better to just substitute for one variable into the equation. This is particularly useful when it comes to higher order polynomials, logarithms, trigonometry, etc...



    Please show me a good example where your method works out and I will perfectly point out at which stage you make a substitution just to make it clear.



    There are various methods for obtaining a point of intersections; ie simultaneous equations, or substitution. It just so happens that substitution is the most common form of finding these intersections and you're allowed to substitute because the variables (in both equations) are equivalent at the points of intersections; therefore you are allowed to replace them.



    Well you better get used to this as you do it throughout the entire A-Level maths and you need to understand WHY you do this, as I have explained above.
    Here is what I mean by my method:
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    So I make the equations equal each other, then take one away to get it all as one equation equal to 0. After that, I factorise it to get the x coordinates where the graphs intersect, and then I substitute those x coordinates in to find the corresponding y value.

    I don't understand why for the exam question i referenced to in the original post, you can substitute one equation (y=6x) into the other but with this, you have to do this long winded method.

    On another note, you said that there are plenty of methods of finding the intersection coordinates. When defining what they are, what is the difference between your (and the mark schemes method) of substituting y=6x into the other equation and working directly from there compared to the method I used with this other question which involves taking one away from the other and factorising, and then substitution at the end.

    Also, how would I know whether I can use your method of just substituting one into the other, or whether I would need to do the way I did ti for the other question where you take one away, factorise it, etc. With the question using my method, why can't you use the method of just simply substituting one equation into the other?

    I am trying to understand how I know whether I need to do my method of making it equal 0, factorising, substituting, etc, or whether I am able to just use your method of simple substitution? In a question, how would I know if I am able to simply substitute it or not, or whether it requires the more complex method that I am used to?
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    could someone quote me the mark scheme. looking at every comment but cant seem to find it. should of gone to specsavers :P
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    (Original post by will'o'wisp)
    You can.

    The person above me is trying to explain how it's easier to substitute. I think.

    You can set both equations equal to 0 and solve.

    I think though it's easier to substitute because it's quicker. I'd rather use a method i know to be quicker and the result x rather than use a longer method and still get the same result.

    Same principle, the 2 lines will intersect somewhere. The x and y co-ordinates will be equal somewhere along the lines. So if you substitute to get rid of y to leave you an equation with just x then you will find the x co-ordinate where both lines intersect. Substitution is just another form of saying these 2 things are equal.
    Ok. When I see a question about where lines/graphs intersect, how do I know whether I can A) Use your simple method of just substituting the simpler equation (y=2x) into the more complex equation (x+2y-13=0), or whether I need to B) Take one away from the other, make it equal to 0, factorise it, find the x values and then substitute each x value into one of the equations to find the corresponding y value.

    How do I know/tell from the question whether I am able to just substitute it directly in like the example in the original post, or whether I have to do the long winded method?
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    (Original post by blobbybill)
    Here is what I mean by my method:

    So I make the equations equal each other, then take one away to get it all as one equation equal to 0. After that, I factorise it to get the x coordinates where the graphs intersect, and then I substitute those x coordinates in to find the corresponding y value.
    From this method, since y=x^3-3x^2-4x and y=6x, you SUBSTITUTE the second equation INTO the first one because the y-values are the same. So you get y=6x=x^3-3x^2-4x \Rightarrow 6x=x^3-3x^2-4x and so on to the answer.

    I don't understand why for the exam question i referenced to in the original post, you can substitute one equation (y=6x) into the other but with this, you have to do this long winded method.
    It's the SAME method...
    On another note, you said that there are plenty of methods of finding the intersection coordinates. When defining what they are, what is the difference between your (and the mark schemes method) of substituting y=6x into the other equation and working directly from there compared to the method I used with this other question which involves taking one away from the other and factorising, and then substitution at the end.
    There is no difference, only the equations are in different forms which makes one way faster than the other. Both ways make up the same substitution.
    Also, how would I know whether I can use your method of just substituting one into the other, or whether I would need to do the way I did ti for the other question where you take one away, factorise it, etc. With the question using my method, why can't you use the method of just simply substituting one equation into the other?
    Again, for intersections you always substitute.
    I am trying to understand how I know whether I need to do my method of making it equal 0, factorising, substituting, etc, or whether I am able to just use your method of simple substitution? In a question, how would I know if I am able to simply substitute it or not, or whether it requires the more complex method that I am used to?
    Once again, it's just substitution in all of these intersection cases.
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    (Original post by blobbybill)
    Ok. When I see a question about where lines/graphs intersect, how do I know whether I can A) Use your simple method of just substituting the simpler equation (y=2x) into the more complex equation (x+2y-13=0), or whether I need to B) Take one away from the other, make it equal to 0, factorise it, find the x values and then substitute each x value into one of the equations to find the corresponding y value.

    How do I know/tell from the question whether I am able to just substitute it directly in like the example in the original post, or whether I have to do the long winded method?
    At what point in the example you have shown are you taking one away from the other, making it equal to 0, factorising it etc?

    From what I see that question uses the exact same method (substitution)


    (Original post by Rajeevan)
    could someone quote me the mark scheme. looking at every comment but cant seem to find it. should of gone to specsavers :P
    There is one in post 1 and one in post 8 but I do not know which one you are referring to
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    (Original post by blobbybill)
    Why do you substitute L2 into L1 in order to find where they intersect? The way I was taught to find points of intersection (with cubic graphs, not a question like this one) was to make the equation equal to each other, take one away from the other and solve from there.

    But I just don't understand why you substitute the equation for L2 (y=2x) into L1 (x+2y-13=0). Can somebody please explain why you do that?Thanks!
    Because both Xs are the same, ditto with the Ys when they intersect. This logic is also applied whe you equate the two equations as you prefer to do. It's simultaneous equations. There are 4 ways of solving simultaneous equations:
    3 algebraically:
    - Adding/Subtracting
    - Equating
    - Substituting
    Or you can measure it on a graph.
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    (Original post by RDKGames)
    From this method, since y=x^3-3x^2-4x and y=6x, you SUBSTITUTE the second equation INTO the first one because the y-values are the same. So you get y=6x=x^3-3x^2-4x \Rightarrow 6x=x^3-3x^2-4x and so on to the answer.



    It's the SAME method...


    There is no difference, only the equations are in different forms which makes one way faster than the other. Both ways make up the same substitution.


    Again, for intersections you always substitute.


    Once again, it's just substitution in all of these intersection cases.
    What do you mean by "The y values are the same" with both equations? One has y=.... and the other has ...+2y...
    There is no difference, only the equations are in different forms which makes one way faster than the other. Both ways make up the same substitution.
    So if the equations are in different forms, you can substitute one equation straight into the other and it is as simple as this exam question (y=2x) and (x+2y-13)?What do you do if both equations are in the form y=..., eg where y=x^3 - 3x^2 -4x and y=6x? Step by step, how do you find the points of intersection then?
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    (Original post by blobbybill)
    What do you mean by "The y values are the same" with both equations? One has y=....
    Yes...
    and the other has ...+2y...
    ...and no.

    ...+2y... that is the y-coordinate in bold and THAT is what you replace, so you'd get 2 multiplied by whatever you're plugging in.
    So if the equations are in different forms, you can substitute one equation straight into the other and it is as simple as this exam question (y=2x) and (x+2y-13)?What do you do if both equations are in the form y=..., eg where y=x^3 - 3x^2 -4x and y=6x? Step by step, how do you find the points of intersection then?
    Well I would simply substitute one of the equations for y into the other because they both are equal to a single value of y.

    I'd get x^3-3x^2-4x=6x and carry on to solve for x.

    ...see how the 'methods' of equating and substituting are one and the same?
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    (Original post by SeanFM)
    At what point in the example you have shown are you taking one away from the other, making it equal to 0, factorising it...?
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    Here, in part B.
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    (Original post by blobbybill)
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    Here, in part B.
    No, it is not doing that. Immediately in the first line y = 6x has been substituted into the other equation.
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    (Original post by SeanFM)
    No, it is not doing that. Immediately in the first line y = 6x has been substituted into the other equation.
    How has it been substituted in? The equations were y=x^2 - 3x^2 -4x, and y=6x. So they made them equal to each other, and then took away 6x from both sides to get it equal to 0, giving x^3 - 3x^2 -10x = 0

    I am so confused right now.
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    (Original post by blobbybill)
    How has it been substituted in? The equations were y=x^2 - 3x^2 -4x, and y=6x. So they made them equal to each other, and then took away 6x from both sides to get it equal to 0, giving x^3 - 3x^2 -10x = 0

    I am so confused right now.
    Sorry, I don't fully understand what you're saying. I appreciate that you've been taught one method and are confused by something that looks different but is actually the same - I think at that point you are confusing yourself. I'm not sure how to help but I would say just look at the two methods side by side and see A. whether they do the same thing and B. whether you can use either whenever you want.
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    (Original post by blobbybill)
    How has it been substituted in? The equations were y=x^2 - 3x^2 -4x, and y=6x. So they made them equal to each other, and then took away 6x from both sides to get it equal to 0, giving x^3 - 3x^2 -10x = 0

    I am so confused right now.
    They are equal to each other BECAUSE the line equation has been SUBSTITUTED for the variable y in the cubic.
 
 
 
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