Question

If the vertex of an equilateral triangle is $(2,3)$ and the equation of the opposite side is $x+y=2$, then the equation of other sides is/are

• 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 $l:x+y=2$


Slope of $BC$ is $-1$, so the inclination angle is
$\theta ={135}^{\circ }$
Therefore, the inclination angle of other sides are
${135}^{\circ }+{60}^{\circ }={195}^{\circ }$ and ${135}^{\circ }-{60}^{\circ }={75}^{\circ }$
So, the slopes are
$m_1=\tan {195}^{\circ }=\tan {15}^{\circ }=2-\sqrt{3}{m}_2=\tan {75}^{\circ }=\cot {15}^{\circ }=2+\sqrt{3}$

Hence, the equation of other sides are
$y-3=(2-\sqrt{3})(x-2)\Rightarrow (2-\sqrt{3})x-y=1-2\sqrt{3}$
And
$y-3=(2+\sqrt{3})(x-2)\Rightarrow (2+\sqrt{3})x-y=1+2\sqrt{3}$