Key Concepts and Formulas

• Translation of a Point: Translating a point $(x, y)$ parallel to a line with direction cosines $(\cos \theta, \sin \theta)$ by a distance $r$ results in a new point $(x + r \cos \theta, y + r \sin \theta)$.
• Slope of a Line: The slope of a line $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$. The slope of a line $y = mx + c$ is $m$.
• Perpendicular Lines: If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1 m_2 = -1$.
• Equation of a Line: The equation of a line passing through $(x_1, y_1)$ with slope $m$ is given by $y - y_1 = m(x - x_1)$.

Step-by-Step Solution

Step 1: Find the slope of the line L and the angle it makes with the x-axis.

The line $L$ is given by $x - y = 4$, or $y = x - 4$. Therefore, the slope of line $L$ is $m = 1$. Let $\theta$ be the angle that the line $L$ makes with the positive x-axis. Then, $\tan \theta = 1$, which means $\theta = \frac{\pi}{4}$. So, $\cos \theta = \frac{1}{\sqrt{2}}$ and $\sin \theta = \frac{1}{\sqrt{2}}$.

Step 2: Determine the coordinates of the translated point Q.

The point (2, 1) is translated by a distance of $2\sqrt{3}$ parallel to the line $L$. The new coordinates are given by: $(x', y') = \left(2 + 2\sqrt{3} \cos \theta, 1 + 2\sqrt{3} \sin \theta\right)$ $(x', y') = \left(2 + 2\sqrt{3} \cdot \frac{1}{\sqrt{2}}, 1 + 2\sqrt{3} \cdot \frac{1}{\sqrt{2}}\right)$ $(x', y') = \left(2 + \sqrt{6}, 1 + \sqrt{6}\right)$

However, the point Q lies in the third quadrant. This means both its x and y coordinates are negative. The translation can occur in two directions along the line $x-y=4$. The coordinates we just calculated are for the first quadrant. We must consider translation in the opposite direction. $(x', y') = \left(2 - 2\sqrt{3} \cos \theta, 1 - 2\sqrt{3} \sin \theta\right)$ $(x', y') = \left(2 - 2\sqrt{3} \cdot \frac{1}{\sqrt{2}}, 1 - 2\sqrt{3} \cdot \frac{1}{\sqrt{2}}\right)$ $(x', y') = \left(2 - \sqrt{6}, 1 - \sqrt{6}\right)$ Since $\sqrt{6} \approx 2.45$, then $2 - \sqrt{6} < 0$ and $1 - \sqrt{6} < 0$. Thus, the coordinates of point Q are $(2 - \sqrt{6}, 1 - \sqrt{6})$.

Step 3: Find the slope of the line perpendicular to L.

The slope of line $L$ is $1$. If $m'$ is the slope of the line perpendicular to $L$, then $m' \cdot 1 = -1$, so $m' = -1$.

Step 4: Find the equation of the line passing through Q and perpendicular to L.

The equation of the line passing through $(2 - \sqrt{6}, 1 - \sqrt{6})$ with slope $-1$ is: $y - (1 - \sqrt{6}) = -1(x - (2 - \sqrt{6}))$ $y - 1 + \sqrt{6} = -x + 2 - \sqrt{6}$ $x + y = 3 - 2\sqrt{6}$ $x + y = 3 - 2\sqrt{6}$

Step 5: Re-examine the problem statement and correct answer.

There appears to be an error in the original problem statement or the provided correct answer. The correct answer derived above is $x + y = 3 - 2\sqrt{6}$, which does not match any of the options provided. Let's re-examine the translation part.

The question states the point (2,1) is translated PARALLEL to the line x-y=4. This means the new point lies on a line that is perpendicular to the line x-y=4. Let's find the equation of the line passing through (2,1) that is perpendicular to x-y=4. The slope of x-y=4 is 1. So the slope of the perpendicular line is -1. The equation of the perpendicular line passing through (2,1) is: y - 1 = -1(x-2) y - 1 = -x + 2 x + y = 3

Now, we need to find the coordinates of the point that is $2\sqrt{3}$ away from (2,1) on the line x+y=3, and lies in the third quadrant. Let the point be (2 - a, 1 + a). Since it's on the perpendicular line, the slope is -1, so the difference in x and y is the same. The distance between (2,1) and (2-a, 1+a) is: $\sqrt{(2 - (2-a))^2 + (1 - (1+a))^2} = 2\sqrt{3}$ $\sqrt{a^2 + a^2} = 2\sqrt{3}$ $\sqrt{2a^2} = 2\sqrt{3}$ $\sqrt{2} |a| = 2\sqrt{3}$ $|a| = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{6}$ Since the point is in the third quadrant, the new point is (2 - $\sqrt{6}$, 1 - $\sqrt{6}$).

We seek the line that passes through (2 - $\sqrt{6}$, 1 - $\sqrt{6}$) with slope 1 (parallel to x - y = 4). y - (1 - $\sqrt{6}$) = 1(x - (2 - $\sqrt{6}$)) y - 1 + $\sqrt{6}$ = x - 2 + $\sqrt{6}$ y = x - 1 x - y = 1

Let's re-read the question carefully. "The equation of the line passing through Q and PERPENDICULAR to L". So, we want the line through (2 - $\sqrt{6}$, 1 - $\sqrt{6}$) with slope -1. y - (1 - $\sqrt{6}$) = -1(x - (2 - $\sqrt{6}$)) y - 1 + $\sqrt{6}$ = -x + 2 - $\sqrt{6}$ x + y = 3 - 2$\sqrt{6}$

Let's re-examine the possible values for a. If a = -$\sqrt{6}$, the new point is (2 + $\sqrt{6}$, 1 - $\sqrt{6}$). This is not in the third quadrant. Thus, the point in the third quadrant is (2 - $\sqrt{6}$, 1 - $\sqrt{6}$). The line passing through this point perpendicular to x - y = 4 has slope -1. y - (1 - $\sqrt{6}$) = -1(x - (2 - $\sqrt{6}$)) y - 1 + $\sqrt{6}$ = -x + 2 - $\sqrt{6}$ x + y = 3 - 2$\sqrt{6}$

There still is no option that matches. Let's examine option A. x + y = 2 - $\sqrt{6}$ Substituting x = 2 - $\sqrt{6}$, we get y = 2 - $\sqrt{6}$ - x = 2 - $\sqrt{6}$ - (2 - $\sqrt{6}$) = 0. This cannot be the correct answer.

Common Mistakes & Tips

• Be careful about the direction of translation. Since the translated point lies in the third quadrant, make sure you are subtracting the correct amounts from the original coordinates.
• Always double-check your calculations to avoid arithmetic errors.
• Read the problem statement carefully and make sure you are answering the question that is asked.

Summary

The point (2, 1) is translated to a new point Q in the third quadrant parallel to the line x - y = 4 by a distance of $2\sqrt{3}$. The coordinates of Q are (2 - $\sqrt{6}$, 1 - $\sqrt{6}$). The equation of the line passing through Q and perpendicular to L is x + y = 3 - 2$\sqrt{6}$. There appears to be an error in the options provided with the original question.

Final Answer

There is an error in the given options. The derived answer is x + y = 3 - 2$\sqrt{6}$. However, if the question had asked for the equation of a line parallel to L, passing through point (1-$\sqrt{6}$, 1), the equation will be x - y = $1 - \sqrt{6} - 1$ = -$\sqrt{6}$. x - y = -$\sqrt{6}$. Also, the options might be incorrect. If the correct answer was x + y = 2 - $\sqrt{6}$, then the equation of the line is incorrect.

The final answer is \boxed{x + y = 3 - 2\sqrt{6}}. There is no correct option.