Here's the question
http://img693.imageshack.us/img693/6272/questionzq.jpg
How would you find the co-ordinates of 'D' in this question?
Thanks!
Printable View
Here's the question
http://img693.imageshack.us/img693/6272/questionzq.jpg
How would you find the co-ordinates of 'D' in this question?
Thanks!
as CD is parallel to AB, shouldn't it be the point C translated to the right 2 and down 4?
Attachment 15296Quote:
Originally Posted by unstopabl3
the slope of AB, according to y1-y2=m(x1-x2) -2, meaning that of BC is its inverse, 1/2. the point C will be where AC an BC cross. find this by setting the equation of each segment equal to one another:
x - 1 = (1/2)x + (11/2), solved gives:
x = 13
so, when plugged into either of the formulae, you will get the point (13,12)
So, the point D is at (11,8)
Can you kindly show me how you got the points of D?Quote:
Originally Posted by NRS
Thanks!
Which software have you used to produce this graph?Quote:
Originally Posted by razemsoft21
Thanks!
This looks suspiciously like an exam question
It surely is ;) Does that make any difference ? :D
Another question is that, if you were not given the equation of AC how would you find that out?
That depends upon your character and the character of those who answer problems by doing all the work without question for others who show little or no effort of their own. It's just a matter of choice.Quote:
Originally Posted by unstopabl3
If it's an exam question that forms part of your graded assessment, then it makes a big difference. MHF policy is to not knowingly help with questions that count towards a student's final grade.Quote:
Originally Posted by unstopabl3
Thread closed.
Edit: The question appears to be from a past exam paper. Thread re-opened.
Thank you Mr Fan ;)
@ Diagonal
I am not the kind to ask someone to do my homework for me, although I don't hesitate to ask for help when required. I think I have followed that ideology in this thread as well. I have not asked you to solve the whole question for me. So please stop bashing.
@ Others
I have asked about something which is not even required in the above question. I am just clearing my concepts. So please do let me know the following two things:
1) How to get the co-ordinates of D
2) If you were not given the equation of AC in the question. Is it possible to find the co-ordinates of C and the equation of AC?
So can someone please elaborate?
Thanks!
Sorry, my mistake perhaps, but this IS the internet of which little girls should be extremely wary, and I can not read minds. To make the point, how would you distinguish honest effort from someone who IS asking to have their work done for them? The easiest way of all is to simply show one's own effort from the start, however right or wrong it may be [and that's the point if one needs useful assistance], and you showed none. I'm not bashing you at all, but have been bashed for having some suspicion that is warranted as I really can not read minds.Quote:
Originally Posted by unstopabl3
Now that that's settled...
Part (i)
BC is perpendicular to AB, so the slope of BC is the negative inverse of that of AB. The slope of AB is (6-2)/(1-3) = -2. The slope of BC is then 1/2. So the equation is (y - 6)/(x - 1) = 1/2. That is, 2y - x = 11.
Part (ii)
You now have two equations to solve for the point of intersection, C. They are...
x - 2y = -11, and x - y = 1
You should be able to solve those to get C(13,12)
Part(iii)
You get from D to C in the same manner as from A to B. They have the same slope. The rise is the same length, as is the run. ...
y - 12 = 2 - 6 So y = 8
x - 13 = 3 - 1 So x = 15
and you have D(15,8)
Now, do you see that I was not "bashing" anyone, but willing to help, and that not I nor anyone can determine honest effort if it is not shown? I am really asking only what is being asked of me ...to show my work.
the software is (Corel Draw) for both graph and solution.Quote:
Originally Posted by unstopabl3
Thanks again.
That's nice, didn't know corel draw had maths related functions/options.Quote:
Originally Posted by razemsoft21
