91Ó°ÊÓ

The plot of the given points $A(1,0),B(5,0)$and $C(7,2)$is:

It is given that the coordinates of the parallelogram ABCD are A(1, 0), B(5, 0) and C(7, 2) .

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 AC is given by:

$\begin{array}{c}O\left(\frac{1+7}{2},\frac{0+2}{2}\right)=O\left(\frac{8}{2},\frac{2}{2}\right)\\ =O\left(4,1\right)\end{array}$

Therefore, the midpoint of AC is $O\left(4,1\right)$.

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

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

$\begin{array}{c}\left(4,1\right)=\left(\frac{5+x}{2},\frac{0+y}{2}\right)\\ \left(4,1\right)=\left(\frac{5+x}{2},\frac{y}{2}\right)\\ ⇒\frac{5+x}{2}=4\text{and}\frac{y}{2}=1\end{array}$

Therefore, it can be obtained that:

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

and

$\begin{array}{c}\frac{y}{2}=1\\ y=2\end{array}$

Therefore, $x=3$and $y=2$.

Therefore, the coordinate of the point D is (3, 2).

Therefore, the coordinate of the fourth vertex is (3, 2).