Question

Two sides of a parallelogram are along the lines, $x+y=3$ and $x-y+3=0$. If its diagonals intersect at $(2,4)$ and one of its vertex is $(a,b)$, then $a+b$ is

Answer 1

Answer: 9

Since, $x-y+3=0$ and $x+y=3$ are perpendicular lines
and intersection point of $x-y+3=0$ and $x+y=3$ is $P(0,3)$.
$\Rightarrow M$ is mid-point of $PR\Rightarrow R(4,5)$
Let $S(x_1,x_2+3)$ and $Q(x_2,3-x_2)$
$M$ is mid-point of $SQ$
$\Rightarrow x_1+x_2=4,x_1+3+3-x_2=8$
$\Rightarrow x_1=3,x_2=1$
Then, the vertex $D$ is $(3,6)$.
$\Rightarrow (a,b)=(3,6)$
$\Rightarrow a=3,b=6$
$\therefore a+b=3+6=9$