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  • The fourth vertex D of a parallelogram ABCD whose three vertices areA (–2, 3), B (6, 7) and C (8, 3) is (A) (0, 1) (B) (0, –1) (C) (–1, 0) (D) (- 2 , 0)

The fourth vertex D of a parallelogram ABCD whose three vertices areA (–2, 3), B (6, 7) and C (8, 3) is

(A) (0, 1)

(B) (0, –1)

(C) (–1, 0)

(D) (- 2 , 0)

Answers (1)

Let D(x, y)

A(–2, 3), B(6, 7), C(8, 3) (given)

We know that in parallelogram diagonals are equal

mid-point of AC = mid-point of BD

\\ \left(\frac{-2+8}{2}, \frac{3+3}{2}\right)=\left(\frac{6+x}{2}, \frac{7+y}{2}\right) \\ \left(\frac{6}{2}, \frac{6}{2}\right)=\left(\frac{6+x}{2}, \frac{7+y}{2}\right) \\ (3,3)=\left(\frac{6+x}{2}, \frac{7+y}{2}\right) \\ \frac{6+x}{2}=3 \quad \frac{7+y}{2}=3 \\ 6+x=6 \quad 7+y=6 \\ x=0 \quad y=-1 \\ \text { Hence, } D=(0,-1)

option B is correct.

Posted by

infoexpert21

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