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

Accepted Answer

Writing the planes as $x-ay-b=0$, $cy-z+d=0$, the line's direction (perpendicular to both normals) is $(a,1,c)$. The point $(a+b,\,1,\,c+d)$ satisfies both plane equations (at $y=1$), giving
$$\frac{x-(a+b)}a=\frac{y-1}1=\frac{z-(c+d)}c.$$

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