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

Answer 1

Answer: (C)

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

Let $(2,1)$ be point A.

$(2,1)$ is at a distance of $2\sqrt{3}$ units from point $Q(a,b).$

Line AQ is parallel to $x-y=4$

Slope of line AQ $=\frac{a-2}{b-1}=1$

$b=a-1$ ----(1)

Also, $(2,1)$ is at a distance of $2\sqrt{3}$ units from point $Q(a,b).$

So, $2\sqrt{3}=\sqrt{(a-2{)}^2+(b-1{)}^2}$

Squaring,
$12=(a-2{)}^2+(b-1{)}^2$

Subsituting equation(1),
$12=(a-2{)}^2+(a-1-1{)}^2$

$12=2(a-2{)}^2$

$a-2=\pm \sqrt{6}$

$a=2\pm \sqrt{6}$

In third quadrant $a=2-\sqrt{6},$
$b=1-\sqrt{6}$

The required line passes through (a,b) and has slope m$=-1,$ since it is perpendicular to L.

So, the equation is $y-(1-\sqrt{6})=-1(x-2+\sqrt{6})$

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