Given:
The three vertices of the parallelogram are (-3,9), (0,-3), (6,-6).
To find:
The fourth vertex of the parallelogram.
Solution:
Consider the three vertices of the parallelogram are A(-3,9), B(0,-3), C(6,-6).
Let D(a,b) be the fourth vertex.
Midpoint formula:
[tex]Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)[/tex]
We know that the diagonals of a parallelogram bisect each other. So, the midpoints of both diagonals are same.
Midpoint of AC = Midpoint BD
[tex]\left(\dfrac{-3+6}{2},\dfrac{9+(-6)}{2}\right)=\left(\dfrac{0+a}{2},\dfrac{-3+b}{2}\right)[/tex]
[tex]\left(\dfrac{3}{2},\dfrac{3}{2}\right)=\left(\dfrac{a}{2},\dfrac{-3+b}{2}\right)[/tex]
On comparing both sides, we get
[tex]\dfrac{3}{2}=\dfrac{a}{2}[/tex]
[tex]3=a[/tex]
And,
[tex]\dfrac{3}{2}=\dfrac{-3+b}{2}[/tex]
[tex]3=-3+b[/tex]
[tex]3+3=b[/tex]
[tex]6=b[/tex]
Therefore, the fourth vertex of the parallelogram is (3,6).
