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Step-by-Step Solution
Step 1: Identify the line L and its slope
The given line is
$x - y = 4$.
We can rewrite it as
$y = x - 4$,
showing that its slope is
$1$.
Step 2: Recognize the perpendicular slope
Any line perpendicular to a line of slope
$m$
will have slope
$- \tfrac{1}{m}$.
Since our line L has slope
$1$,
its perpendicular’s slope is
$-1$.
Step 3: General form of the perpendicular line
A line with slope
$-1$
can be written as
$y = -x + c$.
Rearranging, we get
$x + y - c = 0$.
Step 4: Use the distance condition
We are given that the point
$(2, 1)$
is translated parallel to L by
$2\sqrt{3}$
units to its new location Q, and Q lies on the line
$x + y - c = 0$.
Translating “parallel to L” by
$2\sqrt{3}$
units is equivalent to ensuring the distance from
$(2, 1)$
to the line
$x + y - c = 0$
is
$2\sqrt{3}$.
Step 5: Apply the distance formula
The distance from a point
$(x_1, y_1)$
to the line
$Ax + By + C = 0$
is given by
$ \displaystyle \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Here,
$A = 1,\; B = 1,\; C = -c,\; (x_1, y_1) = (2, 1)$.
The distance is
$2\sqrt{3}$,
so we set up:
$$
\frac{|1\cdot2 + 1\cdot1 - c|}{\sqrt{1^2 + 1^2}} = 2\sqrt{3}.
$$
Simplify inside the absolute value:
$$
\frac{|2 + 1 - c|}{\sqrt{2}} = 2\sqrt{3}
\quad\Longrightarrow\quad
\frac{|3 - c|}{\sqrt{2}} = 2\sqrt{3}.
$$
Multiply both sides by
$\sqrt{2}$:
$$
|3 - c| = 2\sqrt{3}\,\sqrt{2} = 2\sqrt{6}.
$$
Step 6: Solve for c
From
$|3 - c| = 2\sqrt{6}$,
we get two possible values of
$c$:
$$
3 - c = 2\sqrt{6}
\quad\Longrightarrow\quad
c = 3 - 2\sqrt{6},
$$
or
$$
3 - c = -2\sqrt{6}
\quad\Longrightarrow\quad
c = 3 + 2\sqrt{6}.
$$
Hence the two possible perpendicular lines are
$$
x + y = 3 - 2\sqrt{6}
\quad \text{or} \quad
x + y = 3 + 2\sqrt{6}.
$$
Step 7: Determine which line gives Q in the third quadrant
We are told that the new point Q (after translation) must lie in the third quadrant, which means both
$x < 0$
and
$y < 0$.
The line with
$c = 3 + 2\sqrt{6}$
would produce a point Q that is not in the third quadrant for the given distance since it shifts the line further “up/right.”
In contrast,
$c = 3 - 2\sqrt{6}$
(since
$2\sqrt{6}
\approx 4.9$,
$3 - 4.9$
is negative)
leads to coordinates in the third quadrant.
Therefore, the correct line is
$$
x + y = 3 - 2\sqrt{6}.
$$
Final Answer
$x + y = 3 - 2\sqrt{6}.
