Question

If the lines $x=\frac{y}{2}=z,\frac{x-2}{-2}=\frac{y-4}{4}=\frac{z-2}{-1}$ and $4x=\frac{y+h}{1}=\frac{-9z+k}{-2}$ are concurrent, then the value of $(h+k)=$

Answer 1

Given lines : $L_1:x=\frac{y}{2}=z,L_2:\frac{x-2}{-2}=\frac{y-4}{4}=\frac{z-2}{-1}$ $L_3:4x=\frac{y+h}{1}=\frac{-9z+k}{-2}$
Any points on line $L_1=(\lambda ,2\lambda ,\lambda )$ and any points on $L_2=(2-2\mu ,4+4\mu ,2-\mu )$
At point of intersection :
$\lambda =2-2\mu \cdots \left(i\right)2\lambda =4+4\mu \cdots \left(ii\right)\lambda =2-\mu \cdots \left(iii\right)$
Solving $\left(i\right),\left(ii\right):\mu =0,\lambda =2$
So points of intesection is: $(2,4,2)$ lie on line $L_3$ for concurrecny
$\Rightarrow 8=\frac{4+h}{1}=\frac{-18+k}{-2}\Rightarrow h=4,k=2$