Question

A symmetrical form of the line of intersection of the planes $x=ay+b$ and $z=cy+d$ is :

• A. $\frac{x-b}{a}=\frac{y-1}{1}=\frac{z-d}{c}$
• B. $\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$
• C. $\frac{x-a}{b}=\frac{y-0}{1}=\frac{z-c}{d}$
• D. $\frac{x-b-a}{b}=\frac{y-1}{0}=\frac{z-d-c}{d}$

Answer 1

The correct option is B

$\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$

$x-ay-b=0$ ....$\left(1\right)$

$\Rightarrow y=\frac{x-b}{a}$

$-cy+z-d=0$ ....$\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get

$cx=az-ad+bc$ .

$\Rightarrow x=\frac{az-ad+bc}{c}$

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

$\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}$