as CD is parallel to AB, shouldn't it be the point C translated to the right 2 and down 4?
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)
Edit: The question appears to be from a past exam paper. Thread re-opened.
Thank you Mr Fan
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.
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?
Now that that's settled...
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.
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)
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.