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C1- Coordinate Geometry watch

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    Dear Users,

    I found it hard to find the following question.


    Question1
    The line y = x and y = 2x -5 intersect at the point A. Find the equation of the line with gradient 2/5 that passes trough the point A. (Hint: Solve y =x and y = 2x -5)
    I might be able to work it out if I can solve them simultaniously.

    Question 2
    The lines y=4x-10 and y = x-1 intersect at the point T. Find the equation of the line with gradient -2/3 that passes trhough the point T. Write your answer to the form ax + by + x= 0, where a, b and c are integers.
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    (Original post by Limelight31)
    Dear Users,

    I found it hard to find the following question.


    Question1
    The line y = x and y = 2x -5 intersect at the point A. Find the equation of the line with gradient 2/5 that passes trough the point A. (Hint: Solve y =x and y = 2x -5)
    I might be able to work it out if I can solve them simultaniously.

    Question 2
    The lines y=4x-10 and y = x-1 intersect at the point T. Find the equation of the line with gradient -2/3 that passes trhough the point T. Write your answer to the form ax + by + x= 0, where a, b and c are integers.
    What you've said for Q1 is right - so how can you solve those two equations simultaneously, and what do you do with the answer?
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    Q2

    I did this;
    >4x -10 = x -1
    >4x-x=-1+10
    >3x=9
    x=3
    Intercept being of lines is on (3,0)
    And thus y-y1=m(x-x1)
    And thus y-0=-2/3(x-3)
    >y=-2/3x+2
    multiply by 3 for removing fraction
    3y=-2x+6
    somehow the answer is
    2x+3y-12=0
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    (Original post by SeanFM)
    What you've said for Q1 is right - so how can you solve those two equations simultaneously, and what do you do with the answer?
    That is what the book states, that I should simultaniously solve it.

    Not sure where to start as it has no real values.
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    (Original post by Limelight31)
    That is what the book states, that I should simultaniously solve it.

    Not sure where to start as it has no real values.
    Okay. Do you know why you want to solve those two equations?

    And if I said the word substitution, would that help you solve it?
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    (Original post by Limelight31)
    Q2

    I did this;
    >4x -10 = x -1
    >4x-x=-1+10
    >3x=9
    x=3
    Intercept being of lines is on (3,0)
    And thus y-y1=m(x-x1)
    And thus y-0=-2/3(x-3)
    >y=-2/3x+2
    multiply by 3 for removing fraction
    3y=-2x+6
    somehow the answer is
    2x+3y-12=0
    Good job so far

    It was going really well until after the line x=3 - you got y= 0, but how?
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    I could substitute it maybe, This is my working out for that :
    y=x , y= 2x - 5
    ___
    putting all x on one side;(used book info_)
    x=y
    2x=-y-5
    x=-1/2y-5/2y
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    (Original post by SeanFM)
    Good job so far

    It was going really well until after the line x=3 - you got y= 0, but how?
    haha genious so y - 1 would be 0 and y-0 stays y, or would that be 0 aswell "D
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    (Original post by Limelight31)
    haha genious so y - 1 would be 0 and y-0 stays y, or would that be 0 aswell "D
    You're looking for the point T, which will help you answer the question. You have found that the x coordinate of T is 3. Both of the equations are satisfied by the coordinates of T (because T is on both lines). So what is the y coordinate of T?
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    (Original post by SeanFM)
    You're looking for the point T, which will help you answer the question. You have found that the x coordinate of T is 3. Both of the equations are satisfied by the coordinates of T (because T is on both lines). So what is the y coordinate of T?
    so the point of intercept is not 3,0 but would be putting in the x on one of the equations, and thus getting (3,2) mate?
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    (Original post by Limelight31)
    so the point of intercept is not 3,0 but would be putting in the x on one of the equations, and thus getting (3,2) mate?
    Correct - well done

    What is the next step to answer the question?
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    haha thanks mate, the rest is putting in the values in h formula, and it leads to answer, im sttill having some struggle with the Q1.

    Substitution is the word u suggested, after looking up in the book (few chapters back) i came to the idea of bringing all x'és or y'es
    to one site and start from there.
    Let me work it out:
    Y= x
    Y = 2x -5
    using the substitution
    x=y
    2x=-y-5
    x=-1/2y-5/2

    Then substituting right? I dont know what to do man
 
 
 

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