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(B) Let the slope of the required line be $m$. The line passes through $(1, 2)$,so its equation is $(y - 2) = m(x - 1)$.Given that the angle between t

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$3x + y = 5$

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(B) Let the slope of the required line be $m$. The line passes through $(1, 2)$,so its equation is $(y - 2) = m(x - 1)$.Given that the angle between this line and $y = 2x + 1$ is $45^{\circ}$.The slope of the given line is $m_1 = 2$.Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m \cdot m_1} 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 options,$3x + y = 5$ is correct.

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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