Prove that the points A(4,3), B(6,4), C(5,-6) and D(-3,5) are vertices of a parallelogram.

Explanation

Now , distance between A (4,3) and B (6,4), AB `=sqrt((6-4)^(2)+(4-3)^(2))=sqrt(2^(2)+1^(2))=sqrt(5)`
[`:'` distance between the points `(x_(1),y_(1))and(x_(2),y_(2)),d=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))]`
distance between B (6,4) and C(5,-6), BC`=sqrt((5-6)^(2)+(-6-4)^(2))`
`=sqrt((-1)^(2)+(-10)^(2))`
`sqrt(1+100)=sqrt(101)`
Distance between C(5,-6) and D(-3,5),CD `=sqrt((-3-5)^(2)+(5+6)^(2))`
`=sqrt((-8)^(2)+11^(2))`
`=sqrt(64+121)=sqrt(185)`
Distance between D(-3,5) and A (4,3),DA `=sqrt((4+3)^(2)+(3-5)^(2))`
`=sqrt(7^(2)+(-2)^(2))`
`sqrt(49+4)=sqrt(53)`
In parallelogram , opposite sides are equal . Here , we see that all sides AB,BC CD and DA are different .
Henec , given vertices are not the vertices of a parallelogram.