Two sides of a parallelogram are along the lines, and . If its diagonals intersect at , then one of its vertex is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
The two given lines intersect at the vertex: x + y = 3 and x - y + -3 = 0... wait, solving x + y = 3 and x - y = -3 gives 2x = 0, x = 0, y = 3. So one vertex is A = (0, 3).
The diagonals of a parallelogram bisect each other. The center is (2, 4). So the vertex C opposite to A is the reflection of A through (2, 4):
C = (2·2 - 0, 2·4 - 3) = (4, 5).
Now the other two vertices B and D lie on the lines x + y = 3 and x - y + 3 = 0 (one on each, since these are the two sides through A). Let B = (b, 3 - b) on x + y = 3, and D on x - y + 3 = 0, i.e., D = (d, d + 3).
The midpoint of BD must also equal (2, 4):
(b + d)/2 = 2 → b + d = 4
(3 - b + d + 3)/2 = 4 → 6 - b + d = 8 → d - b = 2
Solving: d = 3, b = 1. So B = (1, 2) and D = (3, 6).
D = (3, 6) is one vertex, which is option (a).
The diagonals of a parallelogram bisect each other. The center is (2, 4). So the vertex C opposite to A is the reflection of A through (2, 4):
C = (2·2 - 0, 2·4 - 3) = (4, 5).
Now the other two vertices B and D lie on the lines x + y = 3 and x - y + 3 = 0 (one on each, since these are the two sides through A). Let B = (b, 3 - b) on x + y = 3, and D on x - y + 3 = 0, i.e., D = (d, d + 3).
The midpoint of BD must also equal (2, 4):
(b + d)/2 = 2 → b + d = 4
(3 - b + d + 3)/2 = 4 → 6 - b + d = 8 → d - b = 2
Solving: d = 3, b = 1. So B = (1, 2) and D = (3, 6).
D = (3, 6) is one vertex, which is option (a).
[JEE Main 2019 · 10 Jan · Shift II]