AS Level Mathematics

I

inturgo

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Apr 26, 2020
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Hello,

I'm unsure to how I'd complete (b) part of this following question:
1587916819603.png
I apologise if this is posted in the wrong thread.
Thank you.
 
inturgo said:
Hello,

I'm unsure to how I'd complete (b) part of this following question:
View attachment 18197
I apologise if this is posted in the wrong thread.
Thank you.
Click to expand...
1587917098325.png

Did you calculate k = ??

Please share your calculations.

If I were to do this problem, I would get vector representation of AB, BC, CD and DA

AC = {(7-1), (4-2)} = {6,2} and |AC| = \(\displaystyle \sqrt{40}\)

so, for D to be mid-point of AC, we must have |AD| = 1/2 * |AC| and the unit vectors uAC = uAD

continue......

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.
 
Hello!
Yes, I did get k.
Midpoint of AC:
(1+7)/2 = 4
(2+4)/2 = 3
Therefore making k = 3.
 
IMG_2220.JPG
I think it might be {9,(k-2)^2}.
I'm unsure if I've done this correctly though! ?
 
Of course what you did is not correct. |AD| is a length, that is it is a real number. You seem to have two numbers, 9 and (k-2)^2

Can you please state the distance formula between two points say (x1,y1) and (x2, y2)
And then possible use that formula to find |AD|. Please post back.
 
Thank you for highlighting my mistake Jomo. I figured this problem out and got k = 4 or 0. Substituting these numbers both get to |AD| = sqrt13. I appreciate the help from you & Subhotosh. :)
 
You got the correct results. Good job.
Anytime you are trying to calculate something you need to know what it should look like! If for example it is a distance that you want then it will be a real number (like sqrt(13))or if it is a point in the x-y plane then your answer will be in the form (x,y) or x=?? and y=??
 
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