What is the point of intersection of the diagonals of the trapezium?

Direction: Consider the following for the two (02) items that follow :<br>ABCD is an isosceles trapezium and AB is parallel to DC. Let A(2,3), B(4,3), C(5,1) be the vertices.

  1. A. (3, 7/2)
  2. B. (3, 7/3)
  3. C. (7/2, 2)
  4. D. (5/2, 2)

Correct Answer: B. (3, 7/3)

Explanation

The diagonals are AC and BD, where D is (1,1). The equation of AC (using A(2,3) and C(5,1)) is y - 3 = \frac{1-3}{5-2}(x-2) \implies 2x + 3y = 13. The equation of BD (using B(4,3) and D(1,1)) is y - 3 = \frac{1-3}{1-4}(x-4) \implies 2x - 3y = -1. Adding the two equations yields 4x = 12 \implies x = 3. Substituting x=3 into 2x - 3y = -1 gives 6 - 3y = -1 \implies 3y = 7 \implies y = 7/3. The intersection point is (3, 7/3).

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