Question

If the pair of straight lines \[xy - x - y + 1 = 0\]and the line \[ax + 2y - 3 = 0\] are concurrent, then \[a = \]
1) \[ - 1\]
2) \[0\]
3) \[3\]
4) \[1\]

Answer 1

Hint: since only two equations is given to us so let us split the first equation so as to get 3 equations moreover it is given in the problem that the lines are concurrent therefore by equating determinant of coefficients of lines to zero, we will get the value of a.

Complete step by step answer:
Now let us consider the first equation given in the problem
\[xy - x - y + 1 = 0 - - - \left( 1 \right)\]
Now let us write the above equation in simplified form by taking the common factor and rearranging the terms we can write it as
\[x(y - 1) - (y - 1) = 0\]
\[ \Rightarrow (x - 1)(y - 1) = 0\]
\[ \Rightarrow (x - 1) = 0,(y - 1) = 0\]
Let us consider
The equation of \[L1:(x - 1) = 0\]
The equation of\[L2:(y - 1) = 0\] And
the equation of \[L3:ax + 2y - 3 = 0\]
Since given that the lines are concurrent, therefore we can write
\[\det \left( {\begin{array}{*{20}{c}}
  1&0&{ - 1} \\
  0&1&{ - 1} \\
  a&2&{ - 3}
\end{array}} \right) = 0\]
On expanding we get
\[ \Rightarrow 1( - 3 + 2) - 0(0 +a) - 1(0 - a) = 0\]
On simplification we get
\[ - 1 + a = 0\]
Therefore, \[a = 1\]

So, the correct answer is “Option 4”.

Note: Lines that pass through a single point, on a cartesian plane, are called concurrent lines. Further three lines are said to be concurrent if any one of the lines passes through the point of intersection of the other two lines. The point through which the concurrent lines pass is called the point of concurrency. Or Concurrent-lines A set of lines or curves are said to be concurrent if they all intersect. at the same point.