Question

Find the coordinates of the fourth vertex of a parallelogram whose three vertices taken in order are (x+y, x-y), (2x+y, 2x-y) and (x-y, x+y)

Answer 1

Answer:

D(-y, y)

Step-by-step explanation:

Given

ABCD is a parallelogram with vertices A(x+y, x-y) B(2x+y, 2x-y) C(x-y, x+y)

Let D be (a,b)

We know that if ABCD forms a parallelogram then

Midpoint of AC = Midpoint of BD    ____(1)

Midpoint of (x,y) and (p,q) = [(x+p)/2, (y+q)/2]

A(x+y, x-y), C (x-y, x+y)

Midpoint of AC = {(x+y + x-y)/2, (x-y + x+y)/2}

                         = {2x/2, 2x/2}

                         = {x, x}

B(2x+y, 2x-y), D(a,b)

Midpoint of BD = {(2x+y + a)/2, (2x-y + b)/2}

So

Midpoint of AC = Midpoint of BD

⇒{x, x} = {(2x+y + a)/2, (2x-y + b)/2}

Equating the respective x and y coordinates on both sides

⇒ x = (2x+y+a)/2                       |                 ⇒x = (2x-y+b)/2

⇒2x+y + a = 2x                         |                 ⇒2x-y + b = 2x

⇒y + a = 2x-2x = 0                    |                 ⇒-y + b = 2x-2x = 0

∴a = -y                                        |                 ∴ b = y

∴The fourth vertex D(a,b) = D(-y, y)