Question

The line, $L1$ : $(2-\sqrt{3})x–y+\sqrt{3}=0$ passing through $A(1,2)$ is rotated about $A$ by an angle $\frac{\pi }{2}$ 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 $AC=4$. If the area of the triangle $ABC$ is $\sqrt{8+4\sqrt{3}}$ square units, then the maximum value of $(2+\sqrt{3})h+k$ is

• A. $2\sqrt{3}$
• B. $3\sqrt{3}$
• C. $8+2\sqrt{3}$
• D. $8+3\sqrt{3}$

Answer 1

The correct option is D

$8+3\sqrt{3}$

Given : $L_1\equiv (2-\sqrt{3})x–y+\sqrt{3}=0$

Slope of $L_1=(2-\sqrt{3})$ which is equal to $\tan {15}^{\circ }$

$L_1$ is rotated by $\frac{\pi }{2}$ about $A(1,2)$ to get $L_2$

So, slope of $L_2$ is $\tan (15+90{)}^{\circ }=\tan {105}^{\circ }=-(2+\sqrt{3})$

Hence, equation of $L_2$ is $(2+\sqrt{3})x+y-4-\sqrt{3}=0$

Area of the triangle $ABC=\sqrt{(8+4\sqrt{3}}$

So, $\frac{1}{2}\times AC\times AB=\sqrt{8+4\sqrt{3}}$

$=\frac{1}{2}\times 4\times \frac{|(2+\sqrt{3})h+k-4-\sqrt{3}|}{\sqrt{8+4\sqrt{3}}}=\sqrt{8+4\sqrt{3}}$

$\Rightarrow |(2+\sqrt{3})h+k-4-\sqrt{3}|=4+2\sqrt{3}$

$\Rightarrow {\left(\right(2+\sqrt{3})h+k)}_{max}=8+3\sqrt{3}$