The equation of a line which makes an angle o | vedclass

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(A) Given pair of lines: $xy-x-y+1=0$ $\Rightarrow x(y-1)-1(y-1)=0$ $\Rightarrow (x-1)(y-1)=0$ The lines are $x=1$ (a vertical line) and $y=1$ (a hori

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$x-y=5$

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(A) Given pair of lines: $xy-x-y+1=0$ $\Rightarrow x(y-1)-1(y-1)=0$ $\Rightarrow (x-1)(y-1)=0$ 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 the vertical line $x=1$ is $45^{\circ}$. $	an 45^{\circ} = |\frac{m - \infty}{1 + m \cdot \infty}| = |\frac{1}{m}| = 1 \Rightarrow m = \pm 1$. The angle between the line $y=mx+c$ and the horizontal line $y=1$ is $45^{\circ}$. $	an 45^{\circ} = |\frac{m-0}{1+m \cdot 0}| = |m| = 1 \Rightarrow m = \pm 1$. Since the line must make $45^{\circ}$ with both,we check the slopes. For $m=1$,the line is $y=x+c$ or $x-y+c'=0$. Checking the options,$x-y=5$ has a slope of $1$,which satisfies the condition.

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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