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)

$\frac{9}{2}$

Answer 2

$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z -1}{4} = r$ (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
$3r - 1 -k =8r+2$
$\Rightarrow k=-5r -3 = \frac{15}{2} -3 =\frac{9}{2}$