Question 1.

Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (−2, 8), respectively. Then, the coordinates of the vertex D are

A
(−4, 5)
B
(4, 5)
C
(−3, 4)
D
(0, 11)
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Question Explanation

Text Explanation

In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.

Midpoint of AC = (122,1+82\frac{1-2}{2}, \frac{1+8}{2})

Let the coordinates of vertex D be (x, y)

(x+32,y+42\frac{x+3}{2}, \frac{y+4}{2}) = (122,1+82\frac{1-2}{2}, \frac{1+8}{2})

x = -4 and y = 5

The answer is option D.

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