The mid-point of AC is (a, b). So, option D is correct. In a rectangle, the two diagonals bisect each other at their mid-point.
How to prove that the diagonals of a rectangle bisect each other?
- Find the mid-points of both the diagonals of the rectangle
- Mid-point = ((x1+x2)/2, (y1+y2)/2)
- If the midpoints of both the diagonals are the same then they are said to bisect each other.
- If they are not the same, then they are not bisecting each other.
Calculation:
The given rectangle is ABCD
Its vertices have coordinates as
A - (0, 0)
B - (0, 2a)
C - (2a, 2b)
D - (2a, 0)
The diagonals are AC and BD.
Finding their mid-points:
Mid-point of the diagonal AC = ((0 + 2a)/2 , (0 + 2b)/2)
⇒ (2a/2, 2b/2)
⇒ (a, b) ... (1)
Mid-point of the diagonal BD = ((0 + 2a)/2, (2a+0)/2)
⇒ (2a/2, 2b/2)
⇒ (a, b) ...(2)
From (1) and (2), the midpoints of both the diagonals are equal. So, the diagonals of the rectangle ABCD bisect each other.
Hence, proved.
Therefore, the mid-point of the diagonal AC is (a, b).
Learn more about the diagonals of a rectangle here:
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