- anonymous

ABCD is a square where A is the point (0,2) and C is the point (8,4). AC and BD are diagonals of the square and they intersect at E. Find the coordinates of B and D.

- schrodinger

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- anonymous

I have found out the coordinate of E which is(4,3) and the equation of Bd and length of AE

- amistre64

slope is 1/4?

- amistre64

so perp slope is -4

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## More answers

- anonymous

the ans for the coordinate of B is(5,-1) & D is (3,7)

- anonymous

but i do not know how to derive at it

- amistre64

the slope form a to c aint 1 so its not a square that situated nicely like that is it?

- amistre64

slope from a to c is: 2/8 = 1/4

- anonymous

ABCD is a square in the question

- amistre64

e is a+c/2 for midpoint
8,4
0,2
----
8,6 /2 = 4,3 e is good

- amistre64

y=-4x+4(4)+3
y=-4x+19 is the equation of the line perp to ac and thru 4,3

- amistre64

hmmm, so. ; this laptop doesnt make this easier lol

- amistre64

we need a vector that is the length of ae

- anonymous

its \[\sqrt{17}\] length of ae

- amistre64

8,4
-4,3
----
<4,1>; mag = sqrt(17) that ougt to be fun to play with

- anonymous

i cant find a way to find the coordinate through the length of ae LOL

- amistre64

so the vector representation of say eb is our -4/1 slope

- amistre64

<1,-4> is out vector ... duh lol

- amistre64

form e move 1,-4 and -1,4 to get to the other corners

- anonymous

i was tinking of finding the coordinate through the x-axis difference and y-axis difference but it doesnt seem to work

- amistre64

4,3
1,-4
-----
5,-1
4,3
-1,4
-----
3,7

- anonymous

i get it thx:) alot

- amistre64

took a bit my my brain clicked lol

- Hero

So what were the correct points? For some reason, I'm getting (3,7) and (5,-1), but the question is which one is B and which one is D

- anonymous

Hi im very sorry could you explain what is a vector?

- amistre64

|dw:1327672189805:dw|

- amistre64

a vector can be represented as a directed line segment; something like an arrow with distance and direction defines a vector

- anonymous

how did u arrive at a vector of (-1,4)

- amistre64

|dw:1327672246456:dw|

- amistre64

the slope of a line IS its vector i took the slope of the line from A to C

- amistre64

not drawn correctly in this pic, but same concept nonetheless

- anonymous

gradient AC= 1/4 thus the vector is (-1,4 ) and gradient AB= 4 vector is (-4,1)?

- amistre64

the slope of the line from A to C is:
C (8,4)
-A (0,2)
--------
8, 2 ; slope = 2/8; 1/4
the vector from A to C is then <4 , 1>

- amistre64

a vector is notated the same as a point, except for the < > parts that indicate it as a vector.
it is 4 wide and 1 tall; the components of its gradient (slope)

- anonymous

How did u get
C (8,4)
-A (0,2)
--------
8, 2

- amistre64

its called subtraction

- amistre64

8-0 , 4-2
these are the parts of the slope formula

- amistre64

its just easier for me to do them in this way then to try and sort out what goes where when its already done

- amistre64

but to be clear about the vector from A to C, I was a little off in my explaining.
The vector from A to C is actually <8,2>; half of this is going to be to the midpoint of E, since a midpoint is in the middle
<8,2>
------ = <4,1>, is our vector from A to E
2

- anonymous

i see thank you i will try to understand & do the question first, if I still do not understand could I still look for you? thx::)

- amistre64

i might be around; it might be better to repost a new question, or even this one again so that others have a better chance to see it up on the left and give their views as well :)

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