Three vertices of a parallelogram are $(a + b, a - b), (2 a + b, 2 a - b)$ and $(a - b, a + b)$ taken in a order, the fourth vertex is $(- b, b)$
If true enter $1$ else $0$

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Solution

Let D be the fourth vertex.

$D (x, y)$

Given, $C (a - b, a + b)$

Let P be the mid point of $A (a + b, a - b), B (2 a + b, 2 a - b)$

coordinate of mid point of $A C =$ coordinate of mid point of $B D$

$(\frac{a + b + a - b}{2}, \frac{a - b + a + b}{2}) = (\frac{2 a + b + x}{2}, \frac{2 a - b + y}{2})$

$(a, a) = (\frac{2 a + b + x}{2}, \frac{2 a - b + y}{2})$

by solving the above equation, we get

$\frac{2 a + b + x}{2} = a \Rightarrow x = - b$

$\frac{2 a - b + y}{2} = a \Rightarrow y = b$

$D (x, y) = D (- b, b)$ the fourth vertex.

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