# Co-Ordinate Geometry - Finding Co-Ordinates of a Point

• Feb 7th 2010, 04:22 AM
unstopabl3
Co-Ordinate Geometry - Finding Co-Ordinates of a Point
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!
• Feb 7th 2010, 04:42 AM
NRS
as CD is parallel to AB, shouldn't it be the point C translated to the right 2 and down 4?
• Feb 7th 2010, 09:44 AM
razemsoft21
try to continue ...
Quote:

Originally Posted by unstopabl3
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!

Attachment 15296
• Feb 7th 2010, 10:24 AM
NRS
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)
• Feb 20th 2010, 06:52 AM
unstopabl3
Quote:

Originally Posted by NRS

So, the point D is at (11,8)

Can you kindly show me how you got the points of D?

Thanks!
• Feb 20th 2010, 06:53 AM
unstopabl3
Quote:

Originally Posted by razemsoft21

Which software have you used to produce this graph?
Thanks!
• Feb 20th 2010, 07:01 AM
e^(i*pi)
This looks suspiciously like an exam question
• Feb 20th 2010, 07:38 AM
unstopabl3
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?
• Feb 20th 2010, 12:14 PM
Diagonal
Quote:

Originally Posted by unstopabl3
It surely is ;) Does that make any difference ? :D

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.
• Feb 20th 2010, 01:49 PM
mr fantastic
Quote:

Originally Posted by unstopabl3
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?

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.

Edit: The question appears to be from a past exam paper. Thread re-opened.
• Feb 21st 2010, 04:57 AM
unstopabl3
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?

Thanks!
• Feb 22nd 2010, 10:29 AM
Diagonal
Quote:

Originally Posted by unstopabl3
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.

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.

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.
• Mar 5th 2010, 03:43 PM
razemsoft21
Quote:

Originally Posted by unstopabl3
Which software have you used to produce this graph?
Thanks!

the software is (Corel Draw) for both graph and solution.

Thanks again.
• Mar 5th 2010, 09:46 PM
unstopabl3
Quote:

Originally Posted by razemsoft21
the software is (Corel Draw) for both graph and solution.

Thanks again.

That's nice, didn't know corel draw had maths related functions/options.