શિરોબિંદુુ mathrm{A}(1,2), mathrm{B}(alpha, b | vedclass

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Question

Equation of line passes through point $\mathrm{A}(1,2)$ which makes angle $\frac{\pi}{6}$ from $y=x+4$ is

$ \mathrm{y}-2=\frac{1 \pm 	an \frac{\pi

Vedclass · Accepted Answer

$14$

Vedclass · Answer

Equation of line passes through point $\mathrm{A}(1,2)$ which makes angle $\frac{\pi}{6}$ from $y=x+4$ is

$ \mathrm{y}-2=\frac{1 \pm 	an \frac{\pi}{6}}{1 \mp 	an \frac{\pi}{6}}(\mathrm{x}-1) $

$ \mathrm{y}-2=\frac{\sqrt{3} \pm 1}{\sqrt{3} \mp 1}(\mathrm{x}-1)$

$\oplus$                                                           

$y-2=(2+\sqrt{3})(x-1) $

solve with $ y=x+4 $

$x+2=(2+\sqrt{3}) x-2-\sqrt{3} $

$x=\frac{4+\sqrt{3}}{1+\sqrt{3}}$

$\Theta$

$y-2=(2-\sqrt{3})(x-1)$

solve with $\mathrm{y}=\mathrm{x}+4$

$x+2=(2-\sqrt{3}) x-2+\sqrt{3} $

$x=\frac{4-\sqrt{3}}{1-\sqrt{3}}$

$ \alpha^2+\gamma^2=\left(\frac{4+\sqrt{3}}{1+\sqrt{3}}ight)^2+\left(\frac{4-\sqrt{3}}{1-\sqrt{3}}ight)^2 $

$ \alpha^2+\gamma^2=14$

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