Answer:
The given points are the vertices of a rectangle.
Explanation:
Distance formula
d = √(x₂-x₁)² + (y₂-y₁)²
It is given that, The set of points (4, 3), (7, 3), (4, 0), and (7, 0) .
Let ABCD be the quadrilateral with,
A(4, 3), B(7, 3), C(4, 0), and D(7, 0).
To find each side of ABCD
A(4, 3), B(7, 3), C(4, 0), and D(7, 0).
AB = √(x₂-x₁)² + (y₂-y₁)² = √[(7-4)² + (3-3)²] = 3
BC = √(x₂-x₁)² + (y₂-y₁)² = √[(4-7)² + (0-3)²] = √18 = 3√3
CD = √(x₂-x₁)² + (y₂-y₁)² = √[(7-4)² + (0-0)²] = 3
AD = √(x₂-x₁)² + (y₂-y₁)² = √[(7-4)² + (0-3)]² = √18 = 3√3
To find the diagonals
AC = √(x₂-x₁)² + (y₂-y₁)² = √[(4-4)² + (0-3)² ]= 3
BD = √(x₂-x₁)² + (y₂-y₁)² = √[(7-7)² + (0-3)² ]= 3
Conclusion
From the above result we can see that,
AB = CD ,BC = AD and diagonals AC = BD
Opposite sides are equal and diagonals are equal.
Therefore ABCD is a rectangular.