If the pair of straight lines xy - x - y + 1 | vedclass

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(D) The given equation of the pair of straight lines is $xy - x - y + 1 = 0$. Factoring the expression,we get $(x - 1)(y - 1) = 0$. This implies the t

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$1$

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(D) The given equation of the pair of straight lines is $xy - x - y + 1 = 0$. Factoring the expression,we get $(x - 1)(y - 1) = 0$. This implies the two lines are $x - 1 = 0$ and $y - 1 = 0$. The intersection point of these two lines is $(1, 1)$. Since the three lines are concurrent,the point $(1, 1)$ must satisfy the equation of the third line $ax + 2y - 3 = 0$. Substituting $x = 1$ and $y = 1$ into the equation: $a(1) + 2(1) - 3 = 0$. $a + 2 - 3 = 0$. $a - 1 = 0$,which gives $a = 1$.

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

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