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(B) The given line is $x+y-1=0$,which can be written as $y = -x + 1$. The slope of this line is $m_1 = -1$.Let the slopes of the required lines be $m$

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$xy-x-y+1=0$

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(B) The given line is $x+y-1=0$,which can be written as $y = -x + 1$. The slope of this line is $m_1 = -1$.Let the slopes of the required lines be $m$. Since these lines make an angle of $45^{\circ}$ with the given line,we use the formula $	an 	heta = |\frac{m - m_1}{1 + m m_1}|$.Substituting $	heta = 45^{\circ}$ and $m_1 = -1$:$1 = |\frac{m - (-1)}{1 + m(-1)}| = |\frac{m+1}{1-m}|$.This gives two cases:Case $1$: $\frac{m+1}{1-m} = 1$ $\Rightarrow m+1 = 1-m$ $\Rightarrow 2m = 0$ $\Rightarrow m = 0$.The equation of the line passing through $(1, 1)$ with slope $m=0$ is $y-1 = 0(x-1) \Rightarrow y-1 = 0$.Case $2$: $\frac{m+1}{1-m} = -1$ $\Rightarrow m+1 = -1+m$ $\Rightarrow 1 = -1$,which is impossible. This indicates that the other line is vertical (slope is undefined).The equation of the vertical line passing through $(1, 1)$ is $x-1 = 0$.The combined equation is $(y-1)(x-1) = 0$,which simplifies to $xy - x - y + 1 = 0$.

The combined equation of the straight lines passing through $(1, 1)$ and making an angle of $45^{\circ}$ with the straight line $x+y-1=0$ is

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