Mathematics - Straight Lines Question with Solution | TestHub

Equation of line passes through point $\mathrm{A}(1,2)$ which makes angle $\frac{\pi }{6}$ from $y=x+4$ is $\begin{array}{rl}& y-2=\frac{1\pm \tan \frac{\pi }{6}}{1\mp \tan \frac{\pi }{6}}(x-1)\\ & y-2=\frac{\sqrt{3}\pm 1}{\sqrt{3}\mp 1}(x-1)\\ & \begin{array}{cc}\oplus & \oplus \\ \mathrm{y}-2=(2+\sqrt{3})(\mathrm{x}-1)& \mathrm{y}-2=(2-\sqrt{3})(\mathrm{x}-1)\end{array}\\ & \text{ solve with }\mathrm{y}=\mathrm{x}+4\text{ solve with }\mathrm{y}=\mathrm{x}+4\\ & x+2=(2+\sqrt{3})x-2-\sqrt{3}x+2=(2-\sqrt{3})x-2+\sqrt{3}\\ & x=\frac{4+\sqrt{3}}{1+\sqrt{3}}x=\frac{4-\sqrt{3}}{1-\sqrt{3}}\\ & \end{array}$ $\begin{array}{rl}& {\alpha }^2+{\gamma }^2={\left(\frac{4+\sqrt{3}}{1+\sqrt{3}}\right)}^2+{\left(\frac{4-\sqrt{3}}{1-\sqrt{3}}\right)}^2\\ & {\alpha }^2+{\gamma }^2=14\end{array}$