Question

Given: The coordinates of rhombus WXYZ are W(0, 4b), X(2a, 0), Y(0, −4b), and Z(−2a, 0).

Prove: The segments joining the midpoints of a rhombus form a rectangle.

As part of the proof, find the midpoint of WX

(−a, −2b)


(a, −2b)


(a, 2b)

Answer 1

-a,2b

Explanation:

mid point of WX (say P)=((0+2a)/(2),(4b+0)/(2))=(a,2b)

mid point of XY (say Q)=((2a+0)/(2),(0-4b)/(2)=(a,-2b)

mid point of YZ (say R)=((0-2a)/(2),(-4b+0)/(2)=(-a,-2b)

mid point of WZ (say S)=((0-2a)/(2),(4b+0)/(2)=(-a,2b)

PQ=4b

QR=2a

RS=4b

WS=2a

PQ=RS

QR=WS

PR=QS

hence opposite sides are equal and diagonals are also equal.

so it is a (-a,2b)

Answer 2

Explanation:

mid point of WX (say P)=((0+2a)/(2),(4b+0)/(2))=(a,2b)

mid point of XY (say Q)=((2a+0)/(2),(0-4b)/(2)=(a,-2b)

mid point of YZ (say R)=((0-2a)/(2),(-4b+0)/(2)=(-a,-2b)

mid point of WZ (say S)=((0-2a)/(2),(4b+0)/(2)=(-a,2b)

PQ=4b

QR=2a

RS=4b

WS=2a

PQ=RS

QR=WS

`PR=√((-a-a)^2+(2b+2b)^2) =√(x) =√(4a^2+16b^2) \\QS =√((-a-a)^2+(2b+2b)^2) =√(4a^2+16b^2) \\`

PR=QS

hence opposite sides are equal and diagonals are also equal.

so it is a rectangle.