Coordinate Geometry
Translation of point; perpendicular line equation
Grade Class 12
Question:
The point $(2,1)$ is translated parallel to $L:x-y=4$ by $2\sqrt3$ units. If new point $Q$ is in the third quadrant, equation of line through $Q$ perpendicular to $L$ is
$x+y=2-\sqrt6$... see options
$x+y=3-3\sqrt6$
$x+y=3-2\sqrt6$
$2x+3y=1-\sqrt6$
Step-by-Step Solution
Key Concept: Direction of $L$ (slope 1): unit vector $(1/\sqrt2,1/\sqrt2)$. Move $2\sqrt3$ units: $Q=(2+2\sqrt3/\sqrt2,1+2\sqrt3/\sqrt2)$ or opposite. For $Q$ in third quadrant: $Q=(2-\sqrt6,1-\sqrt6)$. Perpendicular to $L$ (slope $-1$): $y-(1-\sqrt6)=-(x-(2-\sqrt6))\Rightarrow x+y=3-2\sqrt6$.
$x+y=3-2\sqrt6$.
Correct Answer: 3