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

Question

A symmetrical form of the line of intersection of the planes x = ay + b and z = cy + d is : (A) (x-b/a)=(y-1/1)=(z-d/c) (B) (x-b-a/a)=(y-1/1)=(z-d-c/c) (C) (x-a/b)=(y-0/1)=(z-c/d) (D) (x-b-a/b)=(y-1/0)=(z-d-c/d)

Tardigrade Admin · Accepted Answer

(x-b/a)=y=(z-d/c) ⇒ (x-b/a)-1=y-1=(z-d/c)-1 (x-b-a/a)=(y-1/1)=(z-d-c/c)

Q. A symmetrical form of the line of intersection of the planes $x = a y + b$ and $z = cy + d$ is :

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

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

$\frac{x - a}{b} = \frac{y - 0}{1} = \frac{z - c}{d}$

$\frac{x - b - a}{b} = \frac{y - 1}{0} = \frac{z - d - c}{d}$

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

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

ax−b​=1y−1​=cz−d​

ax−b−a​=1y−1​=cz−d−c​

bx−a​=1y−0​=dz−c​

bx−b−a​=0y−1​=dz−d−c​