Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Line : \(x = 2y + 1 = z - 1\) | (I) | Crosses \(xz\) plane at (1, 0, 1) |
| (B) | Line : \(x + 1 = 2y + 1 = z\) | (II) | Crosses \(xz\) plane at (0, 0, 1) |
| (C) | Line : \(x - 1 = 2y = z + 1\) | (III) | Crosses \(xz\) plane at (1, 0, -1) |
| (D) | Line : \(x - 1 = 2y = z - 1\) | (IV) | Crosses \(xz\) plane at (1, 0, 2) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
Every point of the xz plane has y = 0, so put y = 0 in each line and read off x and z.
(A) Line : \(x = 2y + 1 = z - 1\)
This matches (IV) (1, 0, 2).
\(x = 0 + 1 = 1,\quad z - 1 = 1,\quad z = 2\)
(B) Line : \(x + 1 = 2y + 1 = z\)
This matches (II) (0, 0, 1).
\(x + 1 = 1,\quad x = 0,\quad z = 1\)
(C) Line : \(x - 1 = 2y = z + 1\)
This matches (III) (1, 0, −1).
\(x - 1 = 0,\quad x = 1,\quad z + 1 = 0,\quad z = -1\)
(D) Line : \(x - 1 = 2y = z - 1\)
This matches (I) (1, 0, 1).
\(x - 1 = 0,\quad x = 1,\quad z - 1 = 0,\quad z = 1\)
So the matching is (A)-(IV), (B)-(II), (C)-(III), (D)-(I).