Three vertices of parallelogram ABCD are A(0,0), B(5,2), and C(8,5)?

Question

What are the possible locations of Point D

I will choose whoever gives the correct answer as the best answer!

I mean the FIRST person with the correct answer**

Answer 1

D(3,3)

Answer 2

ABCD is a parallelogram ==> opposite sides are parallel
AB || CD and BC||AD
we know paralllel lines have equal slope
the slope between two points (x,y) and (m,n) can be expressed
as (n-y) / (m - x) let point D have coordinates of (x,y)
we can get two equations from the two parallel sides
slope of AD = slope of BC
(y - 0) / (x - 0) = (5 - 2) / (8 - 5) ==> y/x = 3/3 ==>
3y = 3x ==>
y = x 1st equation
and
slope of AB = slope of CD
(2 - 0) / (5 - 0) = (5 - y) / (8 - x) ==> 2/5 = (5 - y) / (8 - x) ==>
2(8 - x) = 5(5 - y) ==>
16 - 2x = 25 - 5y ==>
-9 = 2x - 5y 2nd equation
substitute the value of
y from 1st equation into the 2nd equation ==>
-9 = 2x - 5x ==>
-9 = -3x ==>
3 = x
since y = x from first equation then y = 3 therefore
point D has coordinates (3,3)

Answer 3

Since AD = CD ,then,
x^2+y^2 = (5-2)^2 + (8-5)^2 = 18
Since AD and BC must have the same slope, then
y/x = (5-2)/(8-5)= 1
Thus y=x
Thus x^2 +x^2 = 18
2x^2 = 18
x^2 =9
x=3 or -3. Reject -3 as it leads to an impossible situation.
Thus the point D is at (3,3)

Answer 4

Okay, I think there could be 3.
One is at (0+8-5,0+5-2), or (3,3)

Another at (5+0+8,2+0+5) or (13,7).

And the last at (5-(8+0),2-(5+0)), or (-3,-3).

(Note: this would only be if it didn't matter about the order of ABCD. If it does matter, the (3,3) would be D.)

Answer 5

let D be (x,y)
AD^2=BC^2
x^2+y^2=(5-2)^2+(8-5)^2
x^2+y^2=18
AB^2=CD^2
2^2+5^2=(8-x)^2+(5-y)^2
29=64-16x+x^2+25-10y+y^2
=>x^2+y^2-16x-10y=-60
sub x^2+y^2=18
-16x-10y=-78
16x+10y=78
8x+5y=39
(3,3),(8,-5) could be two possible locations