Question

If lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $x-3=\frac{y-k}{2}$ intersect then the value of $k$ is

• A. $\frac{13}{4}$
• B. $\frac{1}{2}$
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Answer 1

The correct option is A $\frac{13}{4}$

Let a point on the first line be $x=2p+1$,$y=3p-1,z=4p+1$,

i.e.,$(2p+1,3p-1,4p+1)$ where $p$ is any constant real number,

As the lines intersect the same point also lies in the other line too,

Therefore the respective co-ordinates are equal,

as there is no $z$ co-ordinate,the value of $p$ is $4p+1=0⟹p=-\frac{1}{4}$

the point which lies on the other line too is $(\frac{1}{2},-\frac{7}{4},0)$,

Substituting this point in the other line gives,

$-5=-\frac{7}{4}-k⟹k=\frac{13}{4}$