A line passing through the point P(1, 1) and | vedclass

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(C) The slope of the line $x - y = 5$ is $m = 1$. Since the required line passes through $P(1, 1)$ and is parallel to $x - y = 5$,its equation is $y -

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$\sqrt{2}$

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(C) The slope of the line $x - y = 5$ is $m = 1$. Since the required line passes through $P(1, 1)$ and is parallel to $x - y = 5$,its equation is $y - 1 = 1(x - 1)$,which simplifies to $y = x$ or $x - y = 0$. To find the intersection point $Q$,we solve the system of equations: $x - y = 0$ $x + 3y = 2$ Substituting $x = y$ into the second equation: $y + 3y = 2 \implies 4y = 2 \implies y = 0.5$. Thus,$x = 0.5$. The point $Q$ is $(0.5, 0.5)$. The length of segment $PQ$ is $\sqrt{(1 - 0.5)^2 + (1 - 0.5)^2} = \sqrt{(0.5)^2 + (0.5)^2} = \sqrt{0.25 + 0.25} = \sqrt{0.5} = \frac{1}{\sqrt{2}}$. Twice the length of $PQ$ is $2 	imes \frac{1}{\sqrt{2}} = \sqrt{2}$.

$A$ line passing through the point $P(1, 1)$ and parallel to the line $x - y = 5$ cuts the line $x + 3y - 2 = 0$ at $Q$. Then twice the length of the segment $PQ$ is

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