Question

If the pair of straight lines \(xy - x + y - 1 = 0\) and the line \(x + ky - 3 = 0\) are concurrent, then the value of \(k\) is equal to

• A. \(4\)
• B. \(3\)
• C. \(-1\)
• D. \(2\)

Hint:

For concurrency with a pair of lines, first factor the pair, find their intersection point, and then substitute into the third line.

Answer 1

The Correct Option is A

Concept:
If three lines are concurrent, then the third line must pass through the point of intersection of the first two lines. ip

Step 1: Factor the given pair of straight lines.
\[ xy-x+y-1=0 \] Group terms: \[ x(y-1)+(y-1)=0 \] \[ (y-1)(x+1)=0 \] So the pair of lines is: \[ y-1=0 \quad \text{and} \quad x+1=0 \] That is, \[ y=1 \quad \text{and} \quad x=-1 \] ip

Step 2: Find their point of intersection.
The two lines intersect at: \[ (-1,1) \] ip

Step 3: Substitute this point into the third line.
The third line is: \[ x+ky-3=0 \] Substitute \((x,y)=(-1,1)\): \[ -1+k(1)-3=0 \] \[ k-4=0 \] \[ k=4 \] ip Hence, the correct answer is:
\[ \boxed{(A)\ 4} \]