Question Number 38879 by Rio Mike last updated on 30/Jun/18
$${Find}\:{the}\:{equation}\:{of}\:{the}\:{line}\:{through} \\ $$$$\left(\mathrm{2},−\mathrm{3}\right)\:{which}\:{make}\:{angles}\:\mathrm{45}°\:{with} \\ $$$${the}\:{line}\:\mathrm{2}{x}\:−\:{y}\:=\:\mathrm{2}. \\ $$
Answered by MrW3 last updated on 30/Jun/18
$${line}\:\mathrm{2}{x}−{y}=\mathrm{2}: \\ $$$$\mathrm{tan}\:\theta={m}=\mathrm{2} \\ $$$${k}=\mathrm{tan}\:\left(\theta\pm\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{tan}\:\theta\pm\mathrm{tan}\:\frac{\pi}{\mathrm{4}}}{\mathrm{1}\mp\mathrm{tan}\:\theta\:\mathrm{tan}\:\frac{\pi}{\mathrm{4}}}=\frac{{m}\pm\mathrm{1}}{\mathrm{1}\mp{m}} \\ $$$$=\frac{\mathrm{2}\pm\mathrm{1}}{\mathrm{1}\mp\mathrm{2}}=\begin{cases}{−\mathrm{3}}\\{\frac{\mathrm{1}}{\mathrm{3}}}\end{cases} \\ $$$$\Rightarrow{eqn}.\:{of}\:{new}\:{line}: \\ $$$$\frac{{y}+\mathrm{3}}{{x}−\mathrm{2}}=−\mathrm{3}\:{or}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{y}+\mathrm{3}=−\mathrm{3}\left({x}−\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{3}{x}+{y}=\mathrm{3} \\ $$$${or} \\ $$$$\Rightarrow\mathrm{3}\left({y}+\mathrm{3}\right)={x}−\mathrm{2} \\ $$$$\Rightarrow{x}−\mathrm{3}{y}=\mathrm{11} \\ $$