Question

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

• A. $(-1,0)$
• B. $(0,-1)$
• C. $(-1,1)$
• D. none of these

Answer 1

The correct option is A $(-1,0)$

Given,

$P(-6,-4),Q(3,5),R(-2,1),S(\alpha ,\beta )$

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 $PQ=\frac{3-6}{2},\frac{5-4}{2}=\frac{-3}{2},\frac{1}{2}$

midpoint of $RS\Rightarrow \frac{-2+\alpha }{2}=-\frac{3}{2}$

$\Rightarrow \alpha =-3+2=-1$

Now,

$\frac{1+\beta }{2}=\frac{1}{2}$

$\Rightarrow \beta =0$

Therefore, coordinates of 4th vertex is $(-1,0)$