Question

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

• A. $(2,6)$
• B. $(2,1)$
• C. $(3,5)$
• D. $(3,6)$

Answer 1

The correct option is D $(3,6)$

Solving
$x+y=3$

$x-y=-3$

Solving above equations $A(0,3)$

$\frac{x_1+0}{2}=2;x_1=4$ similarly $y_1=5$

$C\Rightarrow (4,5)$

$AB$ and $AD$ are perpendicular (By comparing slopes)

$\therefore$ $ABCD$ is rectangle.

Now equation of $BC$ is $x-y=-1$($BC$ is parallel to $AD$ and passing through $C$)

and equation of $CD$ is $x+y=9$($CD$ is parallel to $AB$ and passing through $C$)

Solving $x+y=9$ and $x-y=-3$

Point $D$ is $(3,6)$