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Q)

A vertex of an equilateral triangle is (1,1) and the opposite side is $x+y+1=0$.Find the equations of other sides?

$\begin {array} {1 1} (A)\;y-1=(2+\sqrt 3)(x-1)\: and\: (y-1)=(2-\sqrt 3)(x-1) \\ (B)\;y-1=(3+\sqrt 2)(x-1)\: and \: y-1=(3-\sqrt 2)(x-1) \\ (C)\;y+1=(2+\sqrt 3)(x+1)\: and \: (y-1)=(2-\sqrt 3)(x-1) \\ (D)\;y+1=(3+\sqrt 2)(x+1)\: and \: (y+1)=(3-\sqrt 2)(x-1) \end {array}$ Comment
A)
Toolbox:
• Slope of the line $ax+by+c=0$ is $-\large\frac{a}{b}$
• The angle between any two lines $(\theta)$ with slopes $m_1$ and $m_2$ is given by $tan\theta=\bigg|\large\frac{m_1-m_2}{1+m_1m_2}\bigg|$
• Eqn. of the line through the point $(x_1),y_1)$ and having slope $m$ is $y-y_1=m(x-x_1)$
Step 1
Let the given vertex be $A(1,1)$
Also given that the equation of one side $BC$ is $x+y+1=0$
$\Rightarrow\:$slope of $BC=m=-1$
Let the slope of $AB=m_1$ and slope of $AC=m_2$
step 2
Since the $\Delta\:ABC$ is equilateral triangle,
angle between the lines $AB$ and $BC$ is $60^{\circ}$ and
angle between $AC$ and $BC$ is also $60^{\circ}$.
We know that angle between any two lines $(\theta)$ with slopes $m_1$ and $m_2$
is given by $tan\theta=\bigg|\large\frac{m_1-m_2}{1+m_1m_2}\bigg|$
$\therefore\:tan60^{\circ}=\bigg|\large\frac{m-m_1}{1+m.m_1}\bigg|$
$\Rightarrow\:\sqrt 3=\bigg|\large\frac{-1-m_1}{1+(-1).m_1}\bigg|$
$\Rightarrow\:\sqrt 3(1-m_1)=-1-m_1$
$\Rightarrow\:m_1(1-\sqrt 3)=-1-\sqrt 3$
$\Rightarrow\:m_1=\large\frac{\sqrt 3+1}{\sqrt 3-1}$
Rationalising the denominator we get $m_1=2+\sqrt 3$
Step 3
Similarly angle between $AC$ and $BC$ is given by
$tan60^{\circ}=\bigg|\large\frac{m_2-m}{1+m.m_2}\bigg|$
$\Rightarrow\:\sqrt 3=\bigg|\large\frac{m_2-(-1)}{1+(-1).m_2}\bigg|$
$\Rightarrow\:\sqrt 3(1-m_2)=1+m_2$
$\Rightarrow\:m_2(1+\sqrt 3)=\sqrt 3-1$
$\Rightarrow\:m_2=\large\frac{\sqrt 3-1}{\sqrt 3+1}$
Rationalising the denominator, we get $m_2=2-\sqrt 3$
Step 4
Hence equation of AB is $y-1=(2+\sqrt 3)(x-1)$
and
Equation of $AC$ is $y-1=(2-\sqrt 3)(x-1)$
Hence (a) is the correct answer.