Points \(A(4,3),\ B(6,4),\ C(5,-6),\ D(-3,5)\) are vertices of a parallelogram.
False.
Step 1: A key property of a parallelogram is that its diagonals bisect each other.
So, if A, B, C, D are vertices of a parallelogram (in some order), then the midpoints of its diagonals must be the same.
Step 2: Assume the natural order \(A\!B\!C\!D\). Then the diagonals are \(AC\) and \(BD\).
Step 3: Find midpoint of diagonal \(AC\).
\(A(4,3),\ C(5,-6)\)
Midpoint formula: \(M = \left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right)\)
\(M_{AC} = \left(\tfrac{4+5}{2},\tfrac{3+(-6)}{2}\right) = \left(\tfrac{9}{2},\tfrac{-3}{2}\right)\)
Step 4: Find midpoint of diagonal \(BD\).
\(B(6,4),\ D(-3,5)\)
\(M_{BD} = \left(\tfrac{6+(-3)}{2},\tfrac{4+5}{2}\right) = \left(\tfrac{3}{2},\tfrac{9}{2}\right)\)
Step 5: Compare the two midpoints.
\(M_{AC} = \left(\tfrac{9}{2},\tfrac{-3}{2}\right)\) and \(M_{BD} = \left(\tfrac{3}{2},\tfrac{9}{2}\right)\).
Since \(\left(\tfrac{9}{2},\tfrac{-3}{2}\right) \ne \left(\tfrac{3}{2},\tfrac{9}{2}\right)\), the diagonals do not bisect each other.
Final Step: Therefore, the given points do not form a parallelogram.
Conclusion: The statement is False.