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)
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)
Question ID - 56927 | SaraNextGen Top Answer
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)
|
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) |
1 Answer
127 votes
Answer Key / Explanation : (d) -
x1 =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)
option (d)
127 votes
Answer Link
127