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)

$x-y=4$
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 $2\sqrt{3}$
$2\sqrt{3}=\frac{|2+1-c|}{\sqrt{2}}$
$2\sqrt{6}=\left|3-c\right|$
$c-3=\pm 2\sqrt{6}c=3\pm 2\sqrt{6}$
Line can $bex+y=3\pm 2\sqrt{6}$
$x+y=3-2\sqrt{6}$