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(A) The given pair of lines is $xy-x-y+1=0$. Factorizing the equation: $x(y-1)-1(y-1)=0$,which gives $(x-1)(y-1)=0$. Thus,the lines are $x=1$ (a verti

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$x-y=5$

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(A) The given pair of lines is $xy-x-y+1=0$. Factorizing the equation: $x(y-1)-1(y-1)=0$,which gives $(x-1)(y-1)=0$. Thus,the lines are $x=1$ (a vertical line) and $y=1$ (a horizontal line). Let the required line be $y=mx+c$. The angle between the line $y=mx+c$ and $y=1$ (slope $m_1=0$) is $45^{\circ}$. Using $	an 	heta = |\frac{m-m_1}{1+mm_1}|$,we get $	an 45^{\circ} = |\frac{m-0}{1+0}| = 1$,so $m=1$ or $m=-1$. If $m=1$,the line is $y=x+c$,or $x-y+c'=0$. Checking the options,$x-y=5$ (or $x-y-5=0$) is a line with slope $m=1$. This line makes an angle of $45^{\circ}$ with both $x=1$ and $y=1$.

The equation of a line which makes an angle of $45^{\circ}$ with each of the pair of lines $xy-x-y+1=0$ is

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