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Since ABCD is a kite, the diagonals are perpendicular and given that BM:MD=1:3 we know that BD bisects AC.
Then M is the midpoint of AC, and the coordinates of M are `((1+3)/2,(3.5+2.5)/2)=(2,3)` .
We can use the distance formula to find BM: `BM=sqrt((3-2)^2+(5-3)^2)=sqrt(5)` . Then `MD=3sqrt(5)` .
D will lie on the line through B and M. The slope of the line is `m=(5-3)/(3-2)=2` . The equation of the line is y-3=2(x-2) or y=2x-1.
If teh coordinates of D are (a,b) then we can write them as (a,2a-1) as D lies on the line BM.
Then we can find the distance from D to M using the point (a,2a-1), while we know that the distance is `3sqrt(5)` ; so
`3sqrt(5)=sqrt(5)sqrt((a-2)^2)` Dividing by `sqrt(5)` we get:
`sqrt((a-2)^2)=3` Now `sqrt(x^2)=|x|` so `sqrt((a-2)^2)=|a-2|`
Then `-(a+2)=3 ==>a=-1`
Point D is at (-1,-3)
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