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

Answer: (C)

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

Answer 2

$x-y=4$ To find equation of R slope of L = 0 is 1 $\Rightarrow$ slope of QR = - 1 Let QR is y = mx + c $y=-x+c$ $x+y-c=0$ distance of QR from (2,1) is $2\sqrt{3}$ $2\sqrt{3}=\frac{\left|2+1-c\right|}{\sqrt{2}}$ $2\sqrt{6}=\left|3-c\right|$ $c-3=\pm2\sqrt{6}\,c=3\pm2\sqrt{6}$ Line can $be x+y=3\pm2\sqrt{6}$ $x+y=3-2\sqrt{6}$