The point `(2,1)` is translated parallel to the line `L: x-y=4` by `2sqrt(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

Options

A. `2x+2y+sqrt(16)-1=0`
B. `x+y=3-3sqrt(6)`
C. `x+y=2-sqrt(6)`
D. `x+y=3-2sqrt(6)`

Answer

D

Explanation

The equation of the line passing through the point (2,1) and parallel to x-y=4 is
`(x-2)/("cos" (pi)/(4))=(y-1)/("sin "(pi)/(4))`
The coordinate of points on this line at a distance of `2sqrt(3)` units from (2,1) are `(x-2)/((1)/(sqrt(2)))=(y-1)/((1)/(sqrt(2)))=pm 2sqrt(3)`
or , `(2=sqrt(6), 1+sqrt(6)) and (2-sqrt(6),1-sqrt(6))`
It is given that Q lies is the third quadrant . So, coordinates of Q are `(2-sqrt(6),1-sqrt(6))`.
The equations of the line passing through Q and perpendicular to x-y=4 is
`y-(1-sqrt(6))=-1(x-(2-sqrt(6)))` or, `x+y=3-2sqrt(6)`