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 :

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)

Tardigrade Admin · Accepted Answer

(x1+0/2) =2; xi = 4 similarly y1=5 C ⇒ (4,5) Now equation of BC is x - y = -1 and equation of CD is x + y = 9 Solving x + y = 9 and x - y = -3 Point D is (3, 6)

Q. 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 :

(2,6)

(2,1)

(3,5)

(3,6)

$\frac{x _{1} + 0}{2} = 2; x_{i} = 4$ similarly $y_{1} = 5$
$C \Rightarrow (4, 5)$
Now equation of $BC$ is $x - y = - 1$
and equation of $C D$ is $x + y = 9$
Solving $x + y = 9$ and $x - y = - 3$
Point $D$ is $(3, 6)$

Q. 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 :