Two sides of a parallelogram are along the lines,$x+y=3\text{ and }x-y+3=0.$If its diagonals intersect

at (2, 4), then one of its vertex is

 

Solution:

Let, equation of $AB\text{ is }x+y=3$                                                 ..(i)

and, equation of $AD\text{ is }x-y=-3$                                            ..(ii)

Solving (i) and (ii), we get $x=0,y=3$

$\therefore$ Coordinates of $A$ are$(0, 3)$.

Now,  $\frac{x_1+0}{2}=2$   

and $\frac{y_1+3}{2}=4$                                                                                            

$\Rightarrow x_1=4\text{ and }{y}_1=5$

$\therefore$ Coordinates of$C$ are (4, 5).

Now, let equation of $BC$ is $x-y=k$

$\therefore$$BC$ passes through $C(4, 5).$ 

$\therefore 4-5=k\Rightarrow k=-1$

 $\therefore$  Equation of$BC$ is  $x-y=-1$                                    …(iii)

Similarly, equation of $CD$ is $x+y=9$                                  …(iv)

Solving (ii) and (iv), we get coordinates of$D$ as (3, 6).
Solving (i) and (iii), we get $B$ as (1, 2).