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If two vertices of a parallelogram are $(3, 2), (- 1, 0)$ and the diagonals cut at $(2, - 5)$ , find the other vertices of the parallelogram.

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(1, - 12)

(5, - 10)

None of these

(1, - 14)

(5, - 11)

(2, - 12)

(5, - 10)

answer is C.

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Detailed Solution

Given that, vertices of a parallelogram are

(3, 2), (- 1, 0)

and the diagonals cut at

(2, - 5)

We need to find the other vertices of the parallelogram.
Drawing the required figure,
Question Image

Midpoint of two point

{(x}_{1}, y_{1})

{(x}_{2}, y_{2})

is,

(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})

.
Since diagonals of a parallelogram bisect each other.
Hence, by mid-point theorem for

AC

(2, - 5) = (\frac{x + 3}{2}, \frac{y + 2}{2})

Equating x coordinates,

\Rightarrow

2 = \frac{x + 3}{2}

\Rightarrow x = 1

And, equating y coordinates,

\Rightarrow

- 5 = \frac{y + 2}{2}

\Rightarrow y = - 12

So, point

C

(1, - 12)

Similarly, for

BD

(2, - 5) = (\frac{u - 1}{2}, \frac{v + 0}{2})

Equating x coordinates,

\Rightarrow

2 = \frac{u - 1}{2}

\Rightarrow u = 5

And, equating y coordinates,

\Rightarrow

- 5 = \frac{v + 0}{2}

\Rightarrow v = - 10

So, point

D

(5, - 10)

.
Thus, the point

C

(1, - 12)

and

D

(5, - 10)

.
Therefore, option (3) is correct.

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