question_answer

If \[(-2,1),\,(a,0),\,(4,b)\] and (1, 2) are the vertices of a parallelogram, then

(i) Find a

(ii) Find b

(iii) Area of the parallelogram.

A)

(i)	(ii)	(iii)
1	\[-1\]	10sq. units

               

B)

(i)	(ii)	(iii)
1	\[-2\]	6 sq. units

               

C)

(i)	(ii)	(iii)
1	1	6 sq. units

               

D)

(i)	(ii)	(iii)
\[-1\]	\[-1\]	6 sq. units

Correct Answer: C

Solution :

Let \[A(-2,1),\,B(a,0),C(4,b)\]and \[D(1,2)\]be the vertices of a parallelogram. Since, diagonals of parallelogram bisect each other.                         \[\therefore \]  Let \[E(x,y)\] be the point where diagonals bisect each other. Hence, coordinates of E from AC \[x=\left( \frac{-2+4}{2} \right)=1\]  ?..(i) and \[y=\frac{1+b}{2}\]   ?..(2) and coordinates of E from BD \[x=\frac{a+1}{2}\] and  \[y=\frac{2+0}{2}=1\]    [from (3)] \[\Rightarrow \] \[\frac{a+1}{2}=1\]  [from (1)] \[\Rightarrow \] \[a=2-1\] \[\Rightarrow \]\[a=1\] and  \[\frac{1+b}{2}=1\,\,\,\Rightarrow b=2-1\,\,\,\Rightarrow \,\,b=1\]                 (i) Value of  \[a=1\] (ii) Value of  \[b=1\] (iii) Value of ABCD = Area of \[\Delta ABC+\] Area of \[\Delta ACD\] Now, area of \[\Delta ABC=3sq.\] units Area of \[\Delta ACD=3sq.\] units Area of parallelogram \[=3+3=6\,sq.\] units.