Question

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

• A. $x +y = 2 - \sqrt{6}$
• B. $x +y = 3 - 3\sqrt{6}$
• C. $x +y = 3 - 2\sqrt{6}$
• D. $2x + 2y = 1 - \sqrt{6}$

Answer 1

The Correct Option is C

To solve the problem, we need to translate the point \((2, 1)\) parallel to the line \(L: x-y = 4\) by \(2\sqrt{3}\) units such that the new point \(Q\) lies in the third quadrant. Then, we need to find the equation of a line through \(Q\) that is perpendicular to line \(L\).

Step 1: Understand the direction of translation. 

The line \(L: x-y=4\) has a slope of 1, indicating that it is oriented at an angle of \(45^\circ\) with the positive x-axis. The direction vector parallel to this line is \((1, 1)\), and the direction vector perpendicular to this direction is \((1, -1)\).

Step 2: Calculate the new point \(Q\).

Since we want to translate the point parallel to line \(L\), the direction vector for translation remains \((1, 1)\), but to ensure \(Q\) is in the third quadrant, we consider reversing the direction: \((-1, -1)\).

The distance of translation is \(2\sqrt{3}\). The unit vector for the direction is \((-1/\sqrt{2}, -1/\sqrt{2})\).

The actual movement of the point is: 

\[(2,1) + 2\sqrt{3}\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) = (2,1) + (-\sqrt{6}, -\sqrt{6}) = (2-\sqrt{6}, 1-\sqrt{6})\]

Step 3: Confirm that point Q is in the third quadrant.

The point \((2-\sqrt{6}, 1-\sqrt{6})\) has both negative x and y coordinates assuming \(\sqrt{6} > 2\), confirming that \(Q\) is in the third quadrant.

Step 4: Determine the line perpendicular to \(L\) through \(Q\).

The perpendicular direction to \(L\) is given by the vector \((1, -1)\). Thus, the equation of the line perpendicular to \(x-y=4\) through point \((2-\sqrt{6}, 1-\sqrt{6})\) is: 

\[y - (1-\sqrt{6}) = -1(x - (2-\sqrt{6})) \] \] \[ y - 1 + \sqrt{6} = -x + 2 - \sqrt{6} \] \] \[ x + y = 3 - 2\sqrt{6}\]

Conclusion: The equation of the line passing through \(Q\) and perpendicular to \(L\) is \((x + y = 3 - 2\sqrt{6})\).