ABCD is a rhombus. Its diagonals `AC and BD` intersect at the point M and satisfy `BD = 2AC`.If he point `D and M` represent the complex numbers `1 + i and 2 - i` respectively, then A represents the complex number......or......

Answer

`3 - (i)/(2) ` or `1 - (3i)/(2)`

Explanation

Given D=(1+i),M =(2-i)
and diagonals of a rhombus bisect each other.
Let `B -= (a + ib)` therefore
`(a+1)/(2)=2,(b+1)/(2)-1`
`rArr a+1 =4,b+1 =-2rArra =3,b=-3`
`a+1 =4, b+1=-2rArra=3,b=-3`
`rArr B=(3-3i)`

Again `DM =sqrt((2-1)^2+(-1-1)^2)=sqrt(1+4)=sqrt(5)`
But `BD=2DM rArr BD=2sqrt(5)`
and `2AC =BD rArr 2 AC = 2sqrt(5)`
`rArr AC = sqrt(5)and AC = 2 AM`
`rArr sqrt(5) = 2AM rArr AM =(sqrt(5))/2`
Now ,let coordinate of A be (x+ iy)
But in a rhombus AD=AB therefore we have
`AD^2 = AB^2`
`rArr (x-1)^2+(y-1)^2=(x-3)^2+(y+3)^2`
`rArr x^2+1-2x+y^2+ 1-2y=x^2+9-6x+y^2+9+6y`
`rArr 4x -8y = 16`
`x-2y=4`
`rArr x=2y+4`
Again `AM =(sqrt(5))/(2)rArr AM^2=(5)/4`
`rArr (x-2)^2 + (y +1)^2 = 5/4 " "["from Eq. (i)"]`
`rArr 5y^2+10 y+5=(5)/4`
`20y^4+40 y+ 15 =0`
`rArr 4y^2 +8y+3=0`
`rArr (2y+1)(2y+3)=0`
`2y+1=0,2y+3 =0`
`rArr y = -1/2,y=-3/2`
on putting these values in Eq.(i) we get
`x=2(-1/2)+4,x=2(-3/2)+4`
`rArr x=3,x=1`
Therefore A is either `(3-i/2) or (1-(3i)/(2))`
Alternate Solution
since, M is the centre of rhombus
`therefore` By rotating D about M through an angle of `pm pi//2` we get possible position of A .

`rArr (z_3-(2-i))/(-1+2 i)=1/2(pmi) rArr (z_3-(2-i))/(-1+2i)=1/2(pmi)`
`rArr z_3 =(2-i)pm1/2i(2i-1)=(2-1)pm1/2(-2-1)`
`=(4-2i -2-i)/(2),(4-2i+2+i)/2=1-3/2i,3-i/2`
`therefore "A is either "(1-3/2i)or(3-i/2)`