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anonymous

  • one year ago

I am getting the wrong answers and I don't know where I am going wrong! Please help! 4. The line with the equation 5x+y=20 meets the x axis at A and the line with the equation x+2y=22 meets the y axis at B. The two lines intersect at point C. (iv) Find the co-ordinates of the point E such that ABEC is a parallelogram. This is what I have done (I don't know if this is right or not): I have found the gradient of BA ( -11/4) which I know will be parallel to EC. I then use the point (0,11) to find the equation of the line BA. I get 4y+11x=44 from this. I do the same for CA (parallel to EB) .

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  1. anonymous
    • one year ago
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    It wouldnt let me type any more so heres the rest

  2. anonymous
    • one year ago
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    I get -5 for the gradient and use point C (2,10) to get the equation of the line 5x+y=20

  3. anonymous
    • one year ago
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    I solve the two equations simultaneously to find where they intersect and only get through the first part till I get -9x=56 which I already know is wrong since x has to equal -2

  4. anonymous
    • one year ago
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    I can post a picture of everything too

  5. Loser66
    • one year ago
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    The first line: 5x +y =20 (1), A on this line and A (4,0) The second line : x+2y = 22 (2) , B on this line and B (0,11) Ok??

  6. anonymous
    • one year ago
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    Yes

  7. anonymous
    • one year ago
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  8. Loser66
    • one year ago
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    |dw:1441547331269:dw|

  9. Loser66
    • one year ago
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    C (2,10) ok?

  10. anonymous
    • one year ago
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    Ok.

  11. Loser66
    • one year ago
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    now, if it forms a parallelogram, then the left side is |dw:1441547541029:dw|

  12. anonymous
    • one year ago
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    Oh wait i think you have A and B wrong.

  13. Loser66
    • one year ago
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    |dw:1441547556565:dw|

  14. Loser66
    • one year ago
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    Show me my mistake, please.

  15. anonymous
    • one year ago
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    This is the answer for the first question it asked saying what the graph looks like

  16. anonymous
    • one year ago
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    So B is on the line on the bottom and is is on the line to the left

  17. Loser66
    • one year ago
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    yes, just estimate. As long as my logic is Ok, I am ok.

  18. anonymous
    • one year ago
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    Ok :)

  19. Loser66
    • one year ago
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    |dw:1441547716564:dw|

  20. Loser66
    • one year ago
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    agree with me for this?

  21. anonymous
    • one year ago
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    Yes, I think so

  22. Loser66
    • one year ago
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    |dw:1441547741256:dw| That is I need a line parallel to 5x+y =20 and passes through B, right?

  23. Loser66
    • one year ago
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    Now, calculate that line by plugging B coordinate in the equation, I get \(5\color{red}{x}+\color{blue}{y}=20\\5\color{red}{0}+\color{blue}{11}=11\), Hence my new line has the equation 5x +y =11|dw:1441547933249:dw|

  24. Loser66
    • one year ago
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    Same as the line parallel to BC which passes through A (4,0)|dw:1441547992236:dw|

  25. Loser66
    • one year ago
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    I have this new line is calculated by \(\color{red}{x}+2\color{blue}{y}=22\\\color{red}{4}+2\color{blue}{0}=4\) hence my brand new line is x +2y =4|dw:1441548113103:dw|

  26. Loser66
    • one year ago
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    So far, I have 2 new lines. Those are 5x+y = 11 and x + 2y =4 Intersection of them is E(2,1)

  27. Loser66
    • one year ago
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    |dw:1441548234109:dw|

  28. Loser66
    • one year ago
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    ok?

  29. anonymous
    • one year ago
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    Um no I dont think thats right... the book says the answer is (-2,21)

  30. Loser66
    • one year ago
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    Whatever the book says, I have a proof of it. https://www.desmos.com/calculator/vku2dwjf3q

  31. Loser66
    • one year ago
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    If the point is -2,21) it can't be form a parallelogram with ABE https://www.desmos.com/calculator/jcjmy55zfm

  32. anonymous
    • one year ago
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    One thing though is that the Book is saying the quadrilateral is ABEC but on the quadrilateral you have shown me A and B don't connect

  33. Loser66
    • one year ago
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    How can they connect? A is intersection point of the first line with the x axis B is intersection point of the second line with the y -axis The two line intersects at C

  34. anonymous
    • one year ago
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    The other thing is, I actually got (2,1) the first time I tried it, but thought it was wrong since he book said so.

  35. Loser66
    • one year ago
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    |dw:1441548556005:dw|

  36. anonymous
    • one year ago
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    I will send a picture of the graph I made

  37. anonymous
    • one year ago
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  38. anonymous
    • one year ago
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    The slightly shaded in area is what I assumed some of the parallelogram would be

  39. anonymous
    • one year ago
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    Sorry, its not the neatest graph in the world

  40. Loser66
    • one year ago
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    Yes, that is the x coordinate of E cannot be negative and y coordinate cannot be exceed 11, right?

  41. anonymous
    • one year ago
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    Ah no? |dw:1441548896467:dw|

  42. anonymous
    • one year ago
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    |dw:1441548990858:dw|

  43. anonymous
    • one year ago
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    And then E would complete it. I know you are saying thats not right, can you explain why?

  44. Loser66
    • one year ago
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    Ok, another proof for this, you have B (0,11) , C (2,10), distance BC is \(BC =\sqrt{(x_C-x_B)^2 +(y_c -y_B)^2}=\sqrt 5\) if A (4,0), E(-2,21) \(AE=\sqrt{477}\) how can they form a parallelogram?

  45. Loser66
    • one year ago
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    That's all I know, if it is wrong still. I am sorry for wasting your time. I am done.

  46. anonymous
    • one year ago
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    Hmm no that makes sense.

  47. anonymous
    • one year ago
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    OK, thank you very much for your help! You didnt waste my time, actually I am sorry for taking so much of yours

  48. anonymous
    • one year ago
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    I hope you have a nice day, for me its back to maths!

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