The point P(2, 1) is translated to a point Q | vedclass

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(D) Given line $L \equiv x-y-4=0$. The slope of line $L$ is $m_1 = 1$.Since the line $PQ$ is parallel to $L$,its slope is also $m_1 = 1$.The equation

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$x+y=3-2\sqrt{6}$

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(D) Given line $L \equiv x-y-4=0$. The slope of line $L$ is $m_1 = 1$.Since the line $PQ$ is parallel to $L$,its slope is also $m_1 = 1$.The equation of line $PQ$ passing through $P(2, 1)$ is $y-1 = 1(x-2)$,which simplifies to $y = x-1$.Let $Q$ be $(x, x-1)$. The distance $PQ = 2\sqrt{3}$.Using the distance formula: $\sqrt{(x-2)^2 + ((x-1)-1)^2} = 2\sqrt{3}$.$\sqrt{(x-2)^2 + (x-2)^2} = 2\sqrt{3} \Rightarrow \sqrt{2(x-2)^2} = 2\sqrt{3}$.$|x-2|\sqrt{2} = 2\sqrt{3} \Rightarrow |x-2| = \sqrt{6}$.So,$x-2 = \sqrt{6}$ or $x-2 = -\sqrt{6}$.$x = 2+\sqrt{6}$ or $x = 2-\sqrt{6}$.Since $Q$ lies in the third quadrant,both coordinates must be negative. For $x = 2+\sqrt{6}$,$y = 1+\sqrt{6}$ (first quadrant). For $x = 2-\sqrt{6}$,$y = (2-\sqrt{6})-1 = 1-\sqrt{6}$ (third quadrant).Thus,$Q = (2-\sqrt{6}, 1-\sqrt{6})$.The line perpendicular to $L$ has slope $m_2 = -1/m_1 = -1$.The equation of the line passing through $Q$ with slope $-1$ is:$y-(1-\sqrt{6}) = -1(x-(2-\sqrt{6}))$.$y-1+\sqrt{6} = -x+2-\sqrt{6}$.$x+y = 3-2\sqrt{6}$.

The point $P(2, 1)$ is translated to a point $Q$ parallel to the line $L \equiv x-y-4=0$ by $2 \sqrt{3}$ units. If the point $Q$ lies in the third quadrant,then the equation of the line passing through $Q$ and perpendicular to $L$ is

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