Answer:
ABCD is a square
Step-by-step explanation:
You want the type of quadrilateral formed by the points A(1, 2), B(4, 5), C(7, 2), and D(4, -1).
Diagonals
The points A and C at either end of diagonal AC both have y-coordinate 2, so lie on the same horizontal line. The distance between them is the difference of their x-coordinates, 7 -1 = 6.
The points B and D at either end of diagonal BD both have x-coordinate 4, so lie on the same vertical line. The distance between them is 5 -(-1) = 6.
The diagonals AC and BD are the same length and perpendicular to each other.
Midpoint
We can check to see if AC and BD have the same midpoint by looking at the sums of their coordinates.
M1 = (A + C)/2 . . . . . . . midpoint of AC
M2 = (B + D)/2 . . . . . . . midpoint of BD
If M1 = M2, we have ...
(A +C)/2 = (B +D)/2
Multiplying by 2 simplifies our test:
A +C = B +D
(1, 2) +(7, 2) = (4, 5) +(4, -1)
(1+7, 2+2) = (4+4, 5-1)
(8, 4) = (8, 4) . . . . . . . . . true.
The diagonals have the same midpoint, so bisect each other.
Conclusion
When the diagonals of a quadrilateral bisect each other, that quadrilateral is a parallelogram. When the diagonals are perpendicular, that parallelogram is a rhombus. When the diagonals of a rhombus are the same length, that rhombus is a square.
The quadrilateral ABCD is a square.
