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. $(3,6)$
• D. $(3,5)$

Answer 1

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

Let the point $C$ is $(x,y)$

$x+y=3\cdots \left(1\right)$
and $x-y+3=0\cdots \left(2\right)$
The point of insection from the equation $\left(1\right)$ and $\left(2\right)$ is point $A:(0,3)$
So, if point $C$ is $(x,y)$ then, $\frac{x+0}{2}=2\Rightarrow x=4$ and
$\frac{y+3}{2}=4\Rightarrow y=5$
As line $BC$ and $AD$ are parrallel
$\therefore$ slope of line $BC=1$
$\therefore$ equation of $BC$ is
$y-5=1(x-4)$
$\Rightarrow x-y+1=0\cdots \left(3\right)$
As line $CD$ and $AB$ are parrallel
$\therefore$ slope of line $CD=-1$
$\therefore$ equation of $CD$ is
$y-5=-1(x-4)$
$\Rightarrow x+y=9\cdots \left(4\right)$
Hence, from $\left(1\right)$ and $\left(3\right)$, point $B$ is $(1,2)$ and from $\left(2\right)$ and $\left(4\right)$, point $D$ is $(3,6)$.