Find the equation of lines which are parallel to each other and at a distance of $4$ units each from line $x = 1$ and forms an equilateral triangle with the line $x = 1$ .

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Solution

Let $O A$ and $O B$ be two lines such that $△ O A B$ is equilateral.

Tangent of equation that is towards positive $x - a x i s$
$m_{1} = t a n 30^{\circ},$
$m_{1} = \frac{1}{\sqrt{3}},$

Equation of line $y = \frac{1}{\sqrt{3}} x + 4$

Or

Equation of line $y - \frac{1}{\sqrt{3}} x - 4 = 0.$

Tangent of equation that is towards negative $x - a x i s$
$m_{2} = - t a n 30^{\circ},$
$m_{2} = \frac{- 1}{\sqrt{3}},$

Equation of line $y = - \frac{1}{\sqrt{3}} x - 4$

Or

$y + \frac{1}{\sqrt{3}} x + 4 = 0.$

Hence,
The joint equation is written as follows
$(y - \frac{1}{\sqrt{3}} x - 4)$ $(y + \frac{1}{\sqrt{3}} x + 4) = 0$

$y^{2} - (\frac{1}{\sqrt{3}} x + 4)^{2} = 0.$

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Find the equation of lines which are parallel to each other and at a distance of 4 units each from line x=1 and forms an equilateral triangle with the line x=1.

Find the equation of lines which are parallel to each other and at a distance of $4$ units each from line $x = 1$ and forms an equilateral triangle with the line $x = 1$ .