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What is the fourth vertex D of a parallelogram ABCD whose three vertices are $A (− 2, 3), B (6, 7)$ and $C (8, 3)$ ?

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(0, - 1)

(1, 2)

(2, 5)

(0, - 3)

answer is A.

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Detailed Solution

It is given that the three vertices of a parallelogram ABCD are

A (− 2, 3), B (6, 7)

and

C (8, 3)

.
We know that the parallelogram's diagonals bisect each other, and the midpoint of the coordinates of both diagonals is equal.
Let, the fourth unknown vertex be (x, y).
Let us draw the figure.
Question Image

By the property of the parallelogram, the midpoint of AC = the midpoint of BD.
The midpoint formula is given by

P (x, y) = (\frac{x_{2} + x_{1}}{2}, \frac{y_{2} + y_{1}}{2})

By placing the values in the above relation, we get the below value.

(− 2 + 8 2, 3 + 3 2) = (6 + x 2, 7 + y 2)

On simplifying the above expression, we get -

(3, 3) = (6 + x 2, 7 + y 2)

On comparing the respective co-ordinates, we get the value of x and y as,

3 = 6 + x 2, 3 = 7 + y 2

⇒ x = 0

and

y = (− 1)

Hence,

(x, y) = (0, − 1)

Since, we get the fourth coordinate

(x, y) = (0, − 1)

.
Hence, the correct option is 1.

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