Question

If the points A (6, 1), B (8, 2), C (9, 4) and D (7, 3) are the vertices of a parallelogram, taken in order, the point of intersection of diagonals will be :

• A. (15/2, 5/2)
• B. (15/2, 5)
• C. (7/2, 5)
• D. (5, 15/2)

Hint:

You only need to find the midpoint of one diagonal (AC or BD) because they share the same intersection point in a parallelogram!

Answer 1

The Correct Option is A

To solve the problem of finding the point of intersection of the diagonals of a parallelogram with vertices A (6, 1), B (8, 2), C (9, 4), and D (7, 3), we can use the property that the diagonals of a parallelogram bisect each other.

The diagonals of a parallelogram bisect each other, meaning the point of intersection of the diagonals is the midpoint of each diagonal. If we denote the diagonals as AC and BD, we need to find the midpoints of these segments.

The midpoint of a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

First, find the midpoint of diagonal AC:

• Coordinates of A: (6, 1)
• Coordinates of C: (9, 4)
• Midpoint of AC: \(\left(\frac{6 + 9}{2}, \frac{1 + 4}{2}\right) = \left(\frac{15}{2}, \frac{5}{2}\right)\)

Next, find the midpoint of diagonal BD:

• Coordinates of B: (8, 2)
• Coordinates of D: (7, 3)
• Midpoint of BD: \(\left(\frac{8 + 7}{2}, \frac{2 + 3}{2}\right) = \left(\frac{15}{2}, \frac{5}{2}\right)\)

Both midpoints calculated are the same, which confirms the diagonals intersect at the point \(\left(\frac{15}{2}, \frac{5}{2}\right)\).

Thus, the correct answer is (15/2, 5/2), which matches with the given option.