anonymous
  • anonymous
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) .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
It wouldnt let me type any more so heres the rest
anonymous
  • anonymous
I get -5 for the gradient and use point C (2,10) to get the equation of the line 5x+y=20
anonymous
  • 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
  • anonymous
I can post a picture of everything too
Loser66
  • 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
  • anonymous
Yes
anonymous
  • anonymous
Loser66
  • Loser66
|dw:1441547331269:dw|
Loser66
  • Loser66
C (2,10) ok?
anonymous
  • anonymous
Ok.
Loser66
  • Loser66
now, if it forms a parallelogram, then the left side is |dw:1441547541029:dw|
anonymous
  • anonymous
Oh wait i think you have A and B wrong.
Loser66
  • Loser66
|dw:1441547556565:dw|
Loser66
  • Loser66
Show me my mistake, please.
anonymous
  • anonymous
This is the answer for the first question it asked saying what the graph looks like
anonymous
  • anonymous
So B is on the line on the bottom and is is on the line to the left
Loser66
  • Loser66
yes, just estimate. As long as my logic is Ok, I am ok.
anonymous
  • anonymous
Ok :)
Loser66
  • Loser66
|dw:1441547716564:dw|
Loser66
  • Loser66
agree with me for this?
anonymous
  • anonymous
Yes, I think so
Loser66
  • Loser66
|dw:1441547741256:dw| That is I need a line parallel to 5x+y =20 and passes through B, right?
Loser66
  • 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
  • Loser66
Same as the line parallel to BC which passes through A (4,0)|dw:1441547992236:dw|
Loser66
  • 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
  • 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
  • Loser66
|dw:1441548234109:dw|
Loser66
  • Loser66
ok?
anonymous
  • anonymous
Um no I dont think thats right... the book says the answer is (-2,21)
Loser66
  • Loser66
Whatever the book says, I have a proof of it. https://www.desmos.com/calculator/vku2dwjf3q
Loser66
  • Loser66
If the point is -2,21) it can't be form a parallelogram with ABE https://www.desmos.com/calculator/jcjmy55zfm
anonymous
  • 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
  • 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
  • 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
  • Loser66
|dw:1441548556005:dw|
anonymous
  • anonymous
I will send a picture of the graph I made
anonymous
  • anonymous
anonymous
  • anonymous
The slightly shaded in area is what I assumed some of the parallelogram would be
anonymous
  • anonymous
Sorry, its not the neatest graph in the world
Loser66
  • Loser66
Yes, that is the x coordinate of E cannot be negative and y coordinate cannot be exceed 11, right?
anonymous
  • anonymous
Ah no? |dw:1441548896467:dw|
anonymous
  • anonymous
|dw:1441548990858:dw|
anonymous
  • anonymous
And then E would complete it. I know you are saying thats not right, can you explain why?
Loser66
  • 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
  • Loser66
That's all I know, if it is wrong still. I am sorry for wasting your time. I am done.
anonymous
  • anonymous
Hmm no that makes sense.
anonymous
  • 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
  • anonymous
I hope you have a nice day, for me its back to maths!

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