Question

Two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=z$ intersect at a point, if $k$ is equal to

• A. $\frac{2}{9}$
• B. $\frac{1}{2}$
• C. $\frac{9}{2}$
• D. $\frac{1}{6}$

Answer 1

Answer: (C)

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=r$ (say)
$\Rightarrow x=2r+1,y=3r-1,z=4r+1$
Since, the two lines intersect.
So, putting above values in second line, we get
$\frac{2r+1-3}{1}=\frac{3r-1-k}{2}=\frac{4r+1}{1}$
Taking 1 st and 3rd terms, we get
$2r-2=4r+1$
$\Rightarrow r=3/2$
Also, taking 2 nd and 3 rd terms, we get
$3r-1-k=8r+2$
$\Rightarrow k=-5r-3=\frac{15}{2}-3=\frac{9}{2}$