# If the lines x=a y+b, z=c y+d

Question:

If the lines $x=a y+b, z=c y+d$ and $x=a^{\prime} z+b^{\prime}$, $\mathrm{y}=\mathrm{c}^{\prime} \mathrm{z}+\mathrm{d}^{\prime}$ are perpendicular, then:

1. $c^{\prime}+a+a^{\prime}=0$

2. $\mathrm{aa}^{\prime}+\mathrm{c}+\mathrm{c}^{\prime}=0$

3. $a b^{\prime}+b c^{\prime}+1=0$

4. $\mathrm{bb}^{\prime}+\mathrm{cc}^{\prime}+1=0$

Correct Option: , 2

Solution:

Line $x=a y+b, z=c y+d \Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}$

Line $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$

$\Rightarrow \frac{\mathrm{x}-\mathrm{b}^{\prime}}{\mathrm{a}^{\prime}}=\frac{\mathrm{y}-\mathrm{d}^{\prime}}{\mathrm{c}^{\prime}}=\frac{\mathrm{z}}{1}$

Given both the lines are perpendicular $\Rightarrow \mathrm{aa}^{\prime}+\mathrm{c}^{\prime}+\mathrm{c}=0$