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If $(- 6, - 4)$ and $(3, 5)$ are the extremities of the diagonals of a parallelogram and $(- 2, 1)$ is its third vertex, then its fourth vertex is

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(- 1, 0)

(0, - 1)

(- 1, 1)

none of these

answer is A.

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Detailed Solution

It is given that

(- 6, - 4)

and

(3, 5)

are the extremities of the diagonals of a parallelogram and

(- 2, 1)

is its third vertex.
Let

P (- 6, - 4), Q (3, 5), R (- 2, 1), S (α, β)

.
Let

P

and

Q

are the extremities of diagonals of a parallelogram, and

R

and

S

will be the extremities of diagonals of a parallelogram.
Now,
Midpoint of

PQ = (\frac{3 - 6}{2}, \frac{5 - 4}{2})

(By the midpoint formula)

= (\frac{- 3}{2}, \frac{1}{2})

Midpoint of

RS

is,

(\frac{- 2 + α}{2}, \frac{1 + β}{2})

(- \frac{3}{2}, \frac{1}{2})

Equating ordinate and abscissa we get,

\frac{- 2 + α}{2}

- \frac{3}{2}

\Rightarrow α - 2 = - 3

\Rightarrow α = - 3 + 2

\Rightarrow α

= - 1

Also,

\frac{1 + β}{2} = \frac{1}{2}

\Rightarrow 1 + β = 1

\Rightarrow β = 1 - 1

\Rightarrow β = 0

So, coordinates of 4 th vertex is

(- 1, 0)

.
So, the correct option is 1.

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