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(B) Let the required equation of the line be $y-2=m(x-1)$.Given that the angle between this line and $y=2x+1$ is $45^{\circ}$.The slope of the given l

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$3x+y=5$

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(B) Let the required equation of the line be $y-2=m(x-1)$.Given that the angle between this line and $y=2x+1$ is $45^{\circ}$.The slope of the given line is $m_1=2$ and the slope of the required line is $m_2=m$.Using the formula $	an 	heta = \left| \frac{m_2-m_1}{1+m_1m_2} ight|$,we have:$	an 45^{\circ} = \left| \frac{m-2}{1+2m} ight|$$1 = \left| \frac{m-2}{1+2m} ight|$This gives two cases:Case $1$: $\frac{m-2}{1+2m} = 1$ $\Rightarrow m-2 = 1+2m$ $\Rightarrow m = -3$.The equation is $y-2 = -3(x-1)$ $\Rightarrow y-2 = -3x+3$ $\Rightarrow 3x+y=5$.Case $2$: $\frac{m-2}{1+2m} = -1$ $\Rightarrow m-2 = -1-2m$ $\Rightarrow 3m = 1$ $\Rightarrow m = \frac{1}{3}$.The equation is $y-2 = \frac{1}{3}(x-1)$ $\Rightarrow 3y-6 = x-1$ $\Rightarrow x-3y+5=0$.Comparing with the given options,$3x+y=5$ is present.

The equation of a straight line passing through the point $(1,2)$ and inclined at $45^{\circ}$ to the line $y=2x+1$ is

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