Question

The symmetrical form of the line $x-y+2z=5,3x+y+z=7$ is

• A. $\frac{4x-11}{-3}=\frac{4y+9}{3}=\frac{z}{1}$
• B. $\frac{x-11}{-3}=\frac{y+9}{3}=\frac{z}{1}$
• C. $\frac{4x-1}{3}=\frac{4y-9}{3}=\frac{z}{1}$
• D. $\frac{x-3}{3}=\frac{y+2}{-5}=\frac{z}{-4}$

Answer 1

The correct option is D $\frac{x-3}{3}=\frac{y+2}{-5}=\frac{z}{-4}$

Putting $z=\lambda ,$

$x-y=5-2\lambda$

$3x+y=7-\lambda$

__________________

$4x=12-3\lambda$

$\Rightarrow x=3-\frac{3}{4}\lambda$

$\Rightarrow \frac{4x-12}{-3}=\lambda$

$y=7-\lambda -3x$

$\Rightarrow y=7-\lambda -9+\frac{9\pi }{4}$

$\Rightarrow y+2=\frac{5\lambda }{4}$

$\Rightarrow \lambda =\frac{4}{5}(y+2)$

$\Rightarrow \frac{4x-12}{-3}=\frac{4y+8}{5}=\frac{z}{1}$

option (A), (B), (C) don't match

For (D), $x=3\lambda +3,y=-5\lambda -2,z=-4\lambda$

putting these $3\lambda +3+5\lambda +2+(-8\lambda )=5$

$\Rightarrow 5=5$

$3(3\lambda +3)+(-5\lambda -2)+(-4\lambda )=7$

$7=7$

$\Rightarrow Option\left(D\right)$