Find the possible co-ordinates of the fourth corner of a parallelogram if its three corners are located at $(3, 3), (4, 4)$, and $(2, 1)$.

Let the corners be $A (3,3) ; B (4,4) ; C (2,1)$

Let the fourth vertex D $= (x,y)$

We know that the diagonals of a parallelogram bisect each other. So, the midpoint of AC is same as the midpoint of BD.

Mid point of two points ${ (x }_{ 1 },{ y }_{ 1 })$ and ${ (x }_{ 2 },{ y }_{ 2 })$ is calculated by the formula $\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right)$.

So, midpoint of $AC =$ Mid point of $BD$

$\Rightarrow \left( \dfrac { 3+2 }{ 2 } ,\dfrac { 3 + 1 }{ 2 } \right) = \left( \dfrac { 4+x }{ 2 } ,\dfrac { 4 +y }{ 2 } \right)$

$\Rightarrow \left( \dfrac { 5 }{ 2 } ,\dfrac { 4 }{ 2 } \right) = \left( \dfrac { 4+x }{ 2 } ,\dfrac { 4 +y }{ 2 } \right)$

$\Rightarrow 4+x=5 ; 4 + y = 4$

$\Rightarrow x = 1 ; y = 0$