91Ó°ÊÓ

Consider the parallelogram be ABCD.

Consider the points $(3,4),(9,4),(6,8)$and $\left(x,y\right)$as A, B, C and D respectively.

In the parallelogram, the diagonals bisect each other.

In the parallelogram ABCD, the diagonals are AC and BD.

Let Obe the point where the diagonals AC and BD intersect each other.

Therefore, the midpoint of AC will be the midpoint of BD that is O is the midpoint of both AC and BD.

The midpoint Y of the line segment joining the point $X\left(x_1,y_1\right)$and $Z\left(x_2,y_2\right)$is given by:

$Y\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Therefore, the midpoint of the line segment AC is given by:

$\begin{array}{c}O\left(\frac{3+6}{2},\frac{4+8}{2}\right)=O\left(\frac{9}{2},\frac{12}{2}\right)\\ =O\left(\frac{9}{2},6\right)\end{array}$

Therefore, the midpoint of AC is $O\left(\frac{9}{2},6\right)$.

As, O is the midpoint of BD, therefore it can be obtained that:

$\begin{array}{c}\left(\frac{9}{2},6\right)=\left(\frac{9+x}{2},\frac{4+y}{2}\right)\\ ⇒\frac{9+x}{2}=\frac{9}{2}\text{and}\frac{4+y}{2}=6\end{array}$

Therefore, it can be obtained that:

$\begin{array}{c}\frac{9+x}{2}=\frac{9}{2}\\ 9+x=9\\ x=9−9\\ x=0\end{array}$

and

$\begin{array}{c}\frac{4+y}{2}=6\\ 4+y=12\\ y=12−4\\ y=8\end{array}$

Therefore, $x=0$and $y=8$.

Therefore, the coordinate of the point D is $\left(0,8\right)$.

Therefore, the coordinate of the fourth vertex is $\left(0,8\right)$.

Consider the parallelogram be ABCD.

Consider the points $(3,4),(9,4),(6,8)$and $\left(x,y\right)$as A, B, C and D respectively.

In the parallelogram, the diagonals bisect each other.

In the parallelogramABCD, the diagonals are AC and AD.

Let O be the point where the diagonals AC and BD intersect each other.

Therefore, the midpoint of AC will be the midpoint of BD that is O is midpoint of both AC and BD.

The midpoint Y of the line segment joining the point $X\left(x_1,y_1\right)$and $Z\left(x_2,y_2\right)$is given by:

$Y\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Therefore, the midpoint of the line segment AC is given by:

$\begin{array}{c}O\left(\frac{3+9}{2},\frac{4+4}{2}\right)=O\left(\frac{12}{2},\frac{8}{2}\right)\\ =O\left(6,4\right)\end{array}$

Therefore, the midpoint of AC is $O\left(6,4\right)$.

As, O is the midpoint of BD, therefore it can be obtained that:

$\begin{array}{c}\left(6,4\right)=\left(\frac{6+x}{2},\frac{8+y}{2}\right)\\ ⇒\frac{6+x}{2}=6\text{and}\frac{8+y}{2}=4\end{array}$

Therefore, it can be obtained that:

$\begin{array}{c}\frac{6+x}{2}=6\\ 6+x=12\\ x=12−6\\ x=6\end{array}$

and

$\begin{array}{c}\frac{8+y}{2}=4\\ 8+y=8\\ y=8−8\\ y=0\end{array}$

Therefore, $x=6$and $y=0$.

Therefore, the coordinate of the point D is $\left(6,0\right)$.

Therefore, the coordinate of the fourth vertex is $\left(6,0\right)$.

Consider the parallelogram be ABCD.

Consider the points $(3,4),(6,8),(x,y)$and $(9,4)$as A, B, C and D respectively.

In the parallelogram, the diagonals bisect each other.

In the parallelogram ABCD, the diagonals are AC and BD.

Let O be the point where the diagonals AC and BD intersect each other.

Therefore, the midpoint of AC will be the midpoint of BD that is O is the midpoint of both AC and BD.

The midpoint Y of the line segment joining the point $X\left(x_1,y_1\right)$and $Z\left(x_2,y_2\right)$is given by:

$Y\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Therefore, the midpoint of the line segment BD is given by:

$\begin{array}{c}O\left(\frac{6+9}{2},\frac{8+4}{2}\right)=O\left(\frac{15}{2},\frac{12}{2}\right)\\ =O\left(\frac{15}{2},6\right)\end{array}$

Therefore, the midpoint of BD is $O\left(\frac{15}{2},6\right)$.

As, O is the midpoint of AC, therefore it can be obtained that:

$\begin{array}{c}\left(\frac{15}{2},6\right)=\left(\frac{3+x}{2},\frac{4+y}{2}\right)\\ ⇒\frac{3+x}{2}=\frac{15}{2}\text{and}\frac{4+y}{2}=6\end{array}$

Therefore, it can be obtained that:

$\begin{array}{c}\frac{3+x}{2}=\frac{15}{2}\\ 3+x=15\\ x=15−3\\ x=12\end{array}$

and

$\begin{array}{c}\frac{4+y}{2}=6\\ 4+y=12\\ y=12−4\\ y=8\end{array}$

Therefore, $x=12$and $y=8$.

Therefore, the coordinate of the point C is $\left(12,8\right)$.

Therefore, the coordinate of the fourth vertex is $\left(12,8\right)$.

Therefore, all possibilities of the fourth vertex are $\left(0,8\right)$, $\left(6,0\right)$and $\left(12,8\right)$.