Given:
The three vertices of a parallelogram are (-3,8), (4,5), (2,-5).
To find:
The fourth vertex of the parallelogram.
Solution:
Let the vertices of the parallelogram are A(-3,8), B(4,5), C(2,-5) and D(a,b).
We know that, diagonals of a parallelogram bisect each other. It means midpoints of both diagonals are same.
Midpoint formula:
![Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)](https://tex.z-dn.net/?f=Midpoint%3D%5Cleft%28%5Cdfrac%7Bx_1%2Bx_2%7D%7B2%7D%2C%5Cdfrac%7By_1%2By_2%7D%7B2%7D%5Cright%29)
Two diagonals of ABCD are AC and BD.
Midpoint of AC = Midpoint of BD
![\left(\dfrac{-3+2}{2},\dfrac{8-5}{2}\right)=\left(\dfrac{4+a}{2},\dfrac{5+b}{2}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B-3%2B2%7D%7B2%7D%2C%5Cdfrac%7B8-5%7D%7B2%7D%5Cright%29%3D%5Cleft%28%5Cdfrac%7B4%2Ba%7D%7B2%7D%2C%5Cdfrac%7B5%2Bb%7D%7B2%7D%5Cright%29)
![\left(\dfrac{-1}{2},\dfrac{3}{2}\right)=\left(\dfrac{4+a}{2},\dfrac{5+b}{2}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B-1%7D%7B2%7D%2C%5Cdfrac%7B3%7D%7B2%7D%5Cright%29%3D%5Cleft%28%5Cdfrac%7B4%2Ba%7D%7B2%7D%2C%5Cdfrac%7B5%2Bb%7D%7B2%7D%5Cright%29)
On comparing both sides, we get
![\dfrac{4+a}{2}=\dfrac{-1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%2Ba%7D%7B2%7D%3D%5Cdfrac%7B-1%7D%7B2%7D)
![4+a=-1](https://tex.z-dn.net/?f=4%2Ba%3D-1)
![a=-1-4](https://tex.z-dn.net/?f=a%3D-1-4)
![a=-5](https://tex.z-dn.net/?f=a%3D-5)
And,
![\dfrac{5+b}{2}=\dfrac{3}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B5%2Bb%7D%7B2%7D%3D%5Cdfrac%7B3%7D%7B2%7D)
![5+b=3](https://tex.z-dn.net/?f=5%2Bb%3D3)
![b=3-5](https://tex.z-dn.net/?f=b%3D3-5)
![b=-2](https://tex.z-dn.net/?f=b%3D-2)
Therefore, the coordinates of fourth vertex are (-5,-2).