Match List-I with List-II
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List-I |
List-II |
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(A) Line: $x = 2y+1=z-1$ |
(1) Crosses $xz$ plane at (1, 0, 1) |
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(B) Line: $x + 1 = 2y+1=z$ |
(II) Crosses $xz$ plane at (0, 0, 1) |
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(C) Line: $x-1= 2y = z+1$ |
(III) Crosses $xz$ plane at (1, 0, -1) |
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(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)
To find where a line crosses the xz-plane, set y = 0 in the line's equation and solve for x and z.
(A) Line: x = 2y + 1 = z − 1
Set y = 0 → x = 1, z = x + 1 = 2 → point = (1, 0, 2)
⇒ Matches with (IV)
(B) Line: x + 1 = 2y + 1 = z
Set y = 0 → 2y + 1 = 1 → x + 1 = 1 ⇒ x = 0, z = 1 → point = (0, 0, 1)
⇒ Matches with (II)
(C) Line: x − 1 = 2y = z + 1
Set y = 0 → 2y = 0 ⇒ x − 1 = 0 ⇒ x = 1, z + 1 = 0 ⇒ z = −1 → point = (1, 0, -1)
⇒ Matches with (III)
(D) Line: x − 1 = 2y = z − 1
Set y = 0 → 2y = 0 ⇒ x − 1 = 0 ⇒ x = 1, z − 1 = 0 ⇒ z = 1 → point = (1, 0, 1)
⇒ Matches with (I)