Question

The equation of the line $x+y+z-1=0$ and $4x+y-2z+2=0$ written in the symmetrical form is:

Answer 1

$x+y+z-1=0$

$4x+y-2z+2=0$

$--+-$

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$-3x+3z-3=0$

$\Rightarrow 3z=3x+3$

$\Rightarrow \frac{z}{1}=\frac{x+1}{1}$

$\Rightarrow \frac{z-1}{1}=\frac{x}{1}..\left(i\right)$

$2x+2y+2z-2=0$

$4x+y-2z+2=0$

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$6x+3y=0$

$\Rightarrow 3y=-6^{2}x$

$\therefore \frac{x}{1}=\frac{y}{-2}...\left(ii\right)$

$\therefore \frac{x}{1}=\frac{y}{-2}=\frac{z-1}{1}$