The line, L 1 : (2−√3)x–y+√3=0 passing through A(1, 2) is rotated about A by an angle π2 in counter clock wise direction to get line L2. Let B(h, k) and C be the points on L1 and L2 respectively such that AC=4. If the area of the triangle ABC is √8+4√3 square units, then the maximum value of (2+√3)h+k is

The line, L 1 : (2−√3)x–y+√3=0 passing through A(1, 2) is rota

1.	The line, L 1 : (2−√3)x–y+√3=0 passing through A(1, 2) is rotated about A by an angle π2 in counter clock wise direction to get line L2. Let B(h, k) and C be the points on L1 and L2 respectively such that AC=4. If the area of the triangle ABC is √8+4√3 square units, then the maximum value of (2+√3)h+k is
Answer» The line, $L 1$ : $(2 - \sqrt{3}) x - y + \sqrt{3} = 0$ passing through $A (1, 2)$ is rotated about $A$ by an angle $\frac{π}{2}$ in counter clock wise direction to get line $L_{2}$ . Let $B (h, k)$ and $C$ be the points on $L_{1}$ and $L_{2}$ respectively such that $A C = 4$ . If the area of the triangle $A B C$ is $\sqrt{8 + 4 \sqrt{3}}$ square units, then the maximum value of $(2 + \sqrt{3}) h + k$ is