Question

The vertex of an equilateral triangle is $(2,3)$ and the opposite side is $x+y=2$ . Then the equations of other sides can be

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

Answer 1

Answer: (A), (C)

The correct options are
A $(2-\sqrt{3})x-y=1-2\sqrt{3}$
C $(2+\sqrt{3})x-y=1+2\sqrt{3}$

Let $A,B$ and $C$ be the vertices of $\Delta ABC$
$A\equiv (2,3)$ and $BC$ lie on the line $l:x+y=2$


Let lines $l_1$ and $l_2$ passing through $A$
$\therefore$ Angle made by $l_1$ and $l_2$ with $l$
i.e. $\alpha$ = ${60}^{\circ }$

Now, $l:x+y=2$
$\Rightarrow$ $y=2-x$
$\Rightarrow$ Slope of $l:\tan \theta =-1$
$\Rightarrow$ $\theta ={180}^{\circ }-{45}^{\circ }={135}^{\circ }$
$\alpha ={60}^{\circ }$ and $\theta ={135}^{\circ }$

$\therefore$ Equations of $l_1$ and $l_2$ are :
$y-3=\tan (\theta \pm \alpha )(x-2)$
$\Rightarrow y-3=\tan ({135}^{\circ }\pm {60}^{\circ })(x-2)$

Case $1:$
$y-3=\tan ({135}^{\circ }+{60}^{\circ })(x-2)$
$\Rightarrow y-3=\tan \left({195}^{\circ }\right)(x-2)$
$\Rightarrow y-3=\tan \left({15}^{\circ }\right)(x-2)$
$\Rightarrow y-3=(2-\sqrt{3})(x-2)$
$\Rightarrow (2-\sqrt{3})x-y=1-2\sqrt{3}$

Case $2:$
$y-3=\tan ({135}^{\circ }-{60}^{\circ })(x-2)$
$\Rightarrow y-3=\tan \left({75}^{\circ }\right)(x-2)$
$\Rightarrow y-3=\cot \left({15}^{\circ }\right)(x-2)$
$\Rightarrow y-3=(2+\sqrt{3})(x-2)$
$\Rightarrow (2+\sqrt{3})x-y=1+2\sqrt{3}$