## 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) .

1. anonymous

It wouldnt let me type any more so heres the rest

2. anonymous

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

3. anonymous

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

I can post a picture of everything too

5. Loser66

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

Yes

7. anonymous

8. Loser66

|dw:1441547331269:dw|

9. Loser66

C (2,10) ok?

10. anonymous

Ok.

11. Loser66

now, if it forms a parallelogram, then the left side is |dw:1441547541029:dw|

12. anonymous

Oh wait i think you have A and B wrong.

13. Loser66

|dw:1441547556565:dw|

14. Loser66

Show me my mistake, please.

15. anonymous

This is the answer for the first question it asked saying what the graph looks like

16. anonymous

So B is on the line on the bottom and is is on the line to the left

17. Loser66

yes, just estimate. As long as my logic is Ok, I am ok.

18. anonymous

Ok :)

19. Loser66

|dw:1441547716564:dw|

20. Loser66

agree with me for this?

21. anonymous

Yes, I think so

22. Loser66

|dw:1441547741256:dw| That is I need a line parallel to 5x+y =20 and passes through B, right?

23. Loser66

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

Same as the line parallel to BC which passes through A (4,0)|dw:1441547992236:dw|

25. Loser66

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

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

|dw:1441548234109:dw|

28. Loser66

ok?

29. anonymous

Um no I dont think thats right... the book says the answer is (-2,21)

30. Loser66

Whatever the book says, I have a proof of it. https://www.desmos.com/calculator/vku2dwjf3q

31. Loser66

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

32. anonymous

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

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

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

|dw:1441548556005:dw|

36. anonymous

I will send a picture of the graph I made

37. anonymous

38. anonymous

The slightly shaded in area is what I assumed some of the parallelogram would be

39. anonymous

Sorry, its not the neatest graph in the world

40. Loser66

Yes, that is the x coordinate of E cannot be negative and y coordinate cannot be exceed 11, right?

41. anonymous

Ah no? |dw:1441548896467:dw|

42. anonymous

|dw:1441548990858:dw|

43. anonymous

And then E would complete it. I know you are saying thats not right, can you explain why?

44. Loser66

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

That's all I know, if it is wrong still. I am sorry for wasting your time. I am done.

46. anonymous

Hmm no that makes sense.

47. anonymous

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

I hope you have a nice day, for me its back to maths!