Question

The point of intersection of the lines $\frac{\left(x-1\right)}{2}=\frac{\left(y-2\right)}{3}=\frac{\left(z-3\right)}{4}$ and $\frac{\left(x-4\right)}{5}=\frac{\left(y-1\right)}{2}=z$ is

• A. $\left(0,0,0\right)$
• B. $\left(1,1,1\right)$
• C. $\left(-1,-1,-1\right)$
• D. $\left(1,2,3\right)$

Answer 1

The correct option is C

$\left(-1,-1,-1\right)$

Explanation for the correct option:

Step 1: Obtain the equation of $x,y$ and $z$.

Find the point of intersection of the given lines

Given: Equation of lines,

$$\begin{gathered} \frac{\left(x-1\right)}{2}=\frac{\left(y-2\right)}{3}=\frac{\left(z-3\right)}{4}\left(1\right) \\ \frac{\left(x-4\right)}{5}=\frac{\left(y-1\right)}{2}=z\left(2\right) \end{gathered}$$

Now, let $\frac{\left(x-1\right)}{2}=\frac{\left(y-2\right)}{3}=\frac{\left(z-3\right)}{4}=r$

$$\begin{gathered} \frac{\left(x-1\right)}{2}=r,\frac{\left(y-2\right)}{3}=r,\frac{z-3}{4}=r \\ \Rightarrow x=2r+1,y=3r+2,z=4r+3 \end{gathered}$$

Step 2: Compute the value of $r$.

Since, $x=2r+1,y=3r+2,z=4r+3$.

Putting these values in equation $\left(2\right)$,

$\frac{\left(2r+1\right)-4}{5}=\frac{\left(3r+2\right)-1}{2}=4r+3$

So,

$$\begin{gathered} \frac{2r+1-4}{5}=4r+3 \\ \Rightarrow 2r-3=20r+15 \\ \Rightarrow -18r=18 \\ \Rightarrow r=-1 \end{gathered}$$

Step 3: Find the point of intersection of the given lines.

Since, $x=2r+1,y=3r+2,z=4r+3$ and $r=-1$.

Substitute $r=-1$ in the equations of $x,y$ and $z$.

So, $x=-1,y=-1,z=-1$.

Therefore, the required point of intersection is $\left(-1,-1,-1\right)$.

Hence, option (C) is the correct answer.