Three vertices of a parallelogram are $(a + b, a – b), (2 a + b, 2a – b), (a – b, a + b)$. Find the fourth vertex.

Given:

Three vertices of a parallelogram are $(a + b, a – b), (2 a + b, 2a – b), (a – b, a + b)$.

To do:

We have to find the fourth vertex.

Solution:

Let the coordinates of the three vertices are $A(a + b, a – b), B(2 a + b, 2a – b), C(a – b, a + b)$.
Let the fourth vertex be $D(x,y)$ and the diagonals $AC$ and $BD$ bisect each other at $O$.

This implies,

\( \mathrm{O} \) is the mid-point of \( \mathrm{AC} \).

The coordinates of \( \mathrm{O} \) are \( (\frac{(a+b)+(a-b)}{2}, \frac{(a-b)+(a+b)}{2}) \)

\( =(\frac{2a}{2}, \frac{2a}{2}) \)

\( =(a,a) \)

\( \mathrm{O} \) is the mid-point of \( \mathrm{BD} \).

The coordinates of \( \mathrm{O} \) are \( (\frac{2a+b+x}{2}, \frac{2a-b+y}{2}) \)

Therefore,

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

On comparing, we get,

\( \frac{2a+b+x}{2}=a \)

\( 2a+b+x=2(a) \)

\( x=2a-2a-b=-b \)

Similarly,

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

\( 2a-b+y=2(a) \)

\( y=2a-2a+b \)

\( y=b \)

Therefore, the coordinates of the fourth vertex are $(-b,b)$.