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) .

- anonymous

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- anonymous

It wouldnt let me type any more so heres the rest

- anonymous

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

- 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

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## More answers

- anonymous

I can post a picture of everything too

- 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??

- anonymous

Yes

- anonymous

##### 1 Attachment

- Loser66

|dw:1441547331269:dw|

- Loser66

C (2,10) ok?

- anonymous

Ok.

- Loser66

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

- anonymous

Oh wait i think you have A and B wrong.

- Loser66

|dw:1441547556565:dw|

- Loser66

Show me my mistake, please.

- anonymous

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

##### 1 Attachment

- anonymous

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

- Loser66

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

- anonymous

Ok :)

- Loser66

|dw:1441547716564:dw|

- Loser66

agree with me for this?

- anonymous

Yes, I think so

- Loser66

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

- 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|

- Loser66

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

- 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|

- Loser66

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

- Loser66

|dw:1441548234109:dw|

- Loser66

ok?

- anonymous

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

- Loser66

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

- Loser66

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

- 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

- 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

- 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.

- Loser66

|dw:1441548556005:dw|

- anonymous

I will send a picture of the graph I made

- anonymous

##### 1 Attachment

- anonymous

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

- anonymous

Sorry, its not the neatest graph in the world

- Loser66

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

- anonymous

Ah no? |dw:1441548896467:dw|

- anonymous

|dw:1441548990858:dw|

- anonymous

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

- 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?

- Loser66

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

- anonymous

Hmm no that makes sense.

- anonymous

OK, thank you very much for your help! You didnt waste my time, actually I am sorry for taking so much of yours

- anonymous

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

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