Question

find the equation of line passing through Point (2, 2 root 3) and inclined with x-axis at the angle of 75 degree

Answer 1

$<b>$As we know that equation of line passing through point(x,y) when slope m is

y -$y_0$ = m (x$-x_0$)

Here point ($x_0$,$y_0$) = (2,2√3)

Hence $x_0$= 2,$y_0$ = 2√3

and slope m= tan Ѳ

Given Ѳ = 75°

m = tan (75°)

= tan(45 + 30)°

$=\frac{ \tan(45°) + \tan(30°) }{ 1 - \tan(45°) \tan(30°) } \\ using \: tan(A+B) = \frac{tan \: A+ tan \: B}{1 - tan A \: \: tan B}$

$= \frac{ 1 + \frac{1}{ \sqrt{3} } }{1 - \frac{1}{\sqrt{3} } }$

$= \frac{ \frac{ \sqrt{3 } + 1 }{ \sqrt{3} } }{ \frac{ \sqrt{3} - 1 }{ \sqrt{ 3} } }$

$\frac{ \sqrt{3} + 1}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3 + 1} } \\ \\ m = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1}$

Putting values in

y - $y_0$ = m(x- $x_0$)

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

$(y - 2 \sqrt{3})( \sqrt{3 } - 1) = (\sqrt{3} + 1)(x - 2)$
$y( \sqrt{3} - 1) - 2 \sqrt{3} ( \sqrt{3} - 1) = x( \sqrt{3} + 1) - 2 \sqrt{3} + 1)$
$y( \sqrt{3} - 1) - 2 \sqrt{3} \times \sqrt{3} + 2 \sqrt{3} = x( \sqrt{3} + 1) - 2 \sqrt{3} - 2)$

$y( \sqrt{3} - 1) - 6 + 2 \sqrt{3} = x( \sqrt{3} + 1) - 2 \sqrt{3} - 1$

$y( \sqrt{3} - 1) = x( \sqrt{3} + 1) - 2\sqrt{3} - 2 + 6 - 2\sqrt{3}$

$y( \sqrt{3} - 1) - = x( \sqrt{3} + 1) - 4\sqrt{3} + 4 \\ \\ y( \sqrt{3} - 1) - x( \sqrt{3} + 1)= - 4 \sqrt{3} + 4$

$y (\sqrt{3} - 1) - x( \sqrt{3} + 1) = 4( - \sqrt{3} + 1) \\ \\ x( \sqrt{3} + 1) - y (\sqrt{3} - 1) = 4(1 - \sqrt{3} )$