Without using distance formula, show that points (-2, -1), (4, 0), (3, 3) and (-3, 2) are vertices of a parallelogram.

Question

A line passing through the point \( A(9, 0) \) makes an angle of \( 30^\circ \) with the positive direction of the x-axis. If this line is rotated about \( A \) through an angle of \( 15^\circ \) in the clockwise direction, then its equation in the new position is:

Answer 1

Given:

Four points are:
A(−2, −1), B(4, 0), C(3, 3) and D(−3, 2)

Step 1: Use the property of diagonals of a parallelogram

A quadrilateral is a parallelogram if its diagonals bisect each other.

So, we find the midpoints of diagonals AC and BD.

Step 2: Find midpoint of diagonal AC

A(−2, −1) and C(3, 3)

Midpoint of AC =

((−2 + 3)/2 , (−1 + 3)/2)

= (1/2 , 1)

Step 3: Find midpoint of diagonal BD

B(4, 0) and D(−3, 2)

Midpoint of BD =

((4 − 3)/2 , (0 + 2)/2)

= (1/2 , 1)

Step 4: Compare the midpoints

Midpoint of AC = Midpoint of BD = (1/2 , 1)

Hence, the diagonals bisect each other.

Conclusion:

Since the diagonals of the quadrilateral bisect each other, the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.

Without using distance formula, show that points (-2, -1), (4, 0), (3, 3) and (-3, 2) are vertices of a parallelogram.

Solution and Explanation

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