91Ó°ÊÓ

The plot of the points $A(3,2),B(8,2)$and $D(0,5)$is:

It is given that the coordinates of the parallelogram ABCD are $A(3,2),B(8,2)$and $D(0,5)$.

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:

$\begin{array}{c}O\left(\frac{8+0}{2},\frac{2+5}{2}\right)=O\left(\frac{8}{2},\frac{7}{2}\right)\\ =O\left(4,\frac{7}{2}\right)\end{array}$

Therefore, the midpoint of BD is $O\left(4,\frac{7}{2}\right)$.

Let the coordinate of point C be $\left(x,y\right)$.

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

$\begin{array}{c}\left(4,\frac{7}{2}\right)=\left(\frac{3+x}{2},\frac{2+y}{2}\right)\\ ⇒\frac{3+x}{2}=4\text{and}\frac{2+y}{2}=\frac{7}{2}\end{array}$

Therefore, it can be obtained that:

$\begin{array}{c}\frac{3+x}{2}=4\\ 3+x=8\\ x=8−3\\ x=5\end{array}$

and

$\begin{array}{c}\frac{2+y}{2}=\frac{7}{2}\\ 2+y=7\\ y=7−2\\ y=5\end{array}$

Therefore, $x=5$and $y=5$.

Therefore, the coordinate of the point C is (5, 5).

Therefore, the coordinate of the fourth vertex is (5, 5).