Match List-I with List-II
| List-I | List-II | ||
| Lines | Direction Ratios | ||
| (A) | \( 2x+3=y+1=z-1 \) | (I) | −1, 2, 0 |
| (B) | \( \displaystyle\frac{1-2x}{2}=\displaystyle\frac{y}{2},z=2 \) | (II) | 1, 1, 1 |
| (C) | \( x=\displaystyle\frac{3y-1}{3}=z+1 \) | (III) | 2, 1, 1 |
| (D) | \( x=2y+3,z=y+1 \) | (IV) | 1, 2, 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
A line written as \( \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \), with coefficient +1 for x, y and z in the numerators, has direction ratios a, b, c.
(A) \( 2x+3=y+1=z-1 \)
This matches (IV) 1, 2, 2.
\( \displaystyle 2x+3=\frac{x+\frac{3}{2}}{\frac{1}{2}} \)
\( \displaystyle \frac{x+\frac{3}{2}}{\frac{1}{2}}=\frac{y+1}{1}=\frac{z-1}{1} \)
Direction ratios \( \frac{1}{2},1,1 \), i.e. 1, 2, 2 (multiplying by 2).
(B) \( \frac{1-2x}{2}=\frac{y}{2},\ z=2 \)
This matches (I) −1, 2, 0.
\( \displaystyle \frac{1-2x}{2}=-\left(x-\frac{1}{2}\right) \)
\( \displaystyle \frac{x-\frac{1}{2}}{-1}=\frac{y}{2} \)
z = 2 is constant along the line, so its direction ratio is 0: direction ratios −1, 2, 0.
(C) \( x=\frac{3y-1}{3}=z+1 \)
This matches (II) 1, 1, 1.
\( \displaystyle \frac{x}{1}=\frac{y-\frac{1}{3}}{1}=\frac{z+1}{1} \)
(D) \( x=2y+3,\ z=y+1 \)
This matches (III) 2, 1, 1.
With y as the common term, \( y=\frac{x-3}{2} \) and \( y=z-1 \):
\( \displaystyle \frac{x-3}{2}=\frac{y}{1}=\frac{z-1}{1} \)
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).