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Question: Answered & Verified by Expert
The two lines $x=a y+b, z=c y+d ;$ and $x=a^{\prime} y+b^{\prime}, z=c^{\prime} y+d^{\prime}$ are perpendicular to each other if
MathematicsStraight LinesJEE MainJEE Main 2006
Options:
  • A
    $\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=-1$
  • B
    $\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=1$
  • C
    $\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}=-1$
  • D
    $\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}=1$
Solution:
1621 Upvotes Verified Answer
The correct answer is:
$\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=-1$
Equation of lines $\frac{x-b}{a}=y=\frac{z-d}{c}$
$\frac{x-b^{\prime}}{a^{\prime}}=y=\frac{z-d^{\prime}}{c^{\prime}}$
Lines are perpendicular $\Rightarrow a a^{\prime}+1+c c^{\prime}=0$

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