The other two vertices are (7, 5) and (8, -1)
Step-by-step explanation:
Given,
Two vertices of a parallelogram are A(1,1) and B(0,7)
The diagonals meet at (4,3)
To find the other two vertices of the parallelogram.
We know that the diagonals of a parallelogram intersect each other.
Let, C be the vertex as (x,y)
According to the problem
![\frac{1+x}{2} = 4 and \frac{1+y}{2} =3](https://tex.z-dn.net/?f=%5Cfrac%7B1%2Bx%7D%7B2%7D%20%20%3D%204%20%20%20%20%20and%20%20%20%5Cfrac%7B1%2By%7D%7B2%7D%20%3D3)
or, 1+x = 8 and 1+y = 6
or, x = 7 and y = 5
Again, let D be the vertex as (a,b)
According to the problem,
![\frac{0+a}{2} = 4 and \frac{7+b}{2} = 3](https://tex.z-dn.net/?f=%5Cfrac%7B0%2Ba%7D%7B2%7D%20%3D%204%20and%20%5Cfrac%7B7%2Bb%7D%7B2%7D%20%3D%203)
or, a = 8 and 7+b = 6
or, a= 8 and b= -1
Hence the vertices are (7, 5) and (8, -1)