Answer:
The given points are the vertices of a rectangle.
Step-by-step explanation:
<u>Distance formula</u>
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).
<u>To find each side of ABCD</u>
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
<u>To find the diagonals</u>
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
<u>Conclusion</u>
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.
