Question Number 150738 by nadovic last updated on 15/Aug/21
$$\mathrm{A}\:\mathrm{point}\:\mathrm{P}\left({x},\:{y}\right)\:\mathrm{moves}\:\mathrm{such}\:\mathrm{that}\:\mathrm{its} \\ $$$$\mathrm{perpendicular}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{12}{x}\:+\:\mathrm{5}{y}\:−\:\mathrm{1}\:=\:\:\mathrm{0}\:\mathrm{is}\:\mathrm{always}\:\mathrm{3}\:{units}.\: \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{that}\:\mathrm{describes}\:\mathrm{the}\: \\ $$$$\mathrm{locus}\:\mathrm{precisely}. \\ $$
Commented by nadovic last updated on 15/Aug/21
$${Thanks}\:{Sir} \\ $$
Commented by JDamian last updated on 15/Aug/21
In other words, find the (two) parallel lines whose distances from the given line are 3 u.
Answered by nimnim last updated on 15/Aug/21
$${y}=−\frac{\mathrm{12}}{\mathrm{5}}{x}+\frac{\mathrm{1}}{\mathrm{5}}\Rightarrow{m}=−\frac{\mathrm{12}}{\mathrm{5}},\:{C}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${Let}\:{the}\:{equation}\:{be}\:{y}={mx}+{C}_{\mathrm{2}} ……\left({i}\right) \\ $$$$\therefore\:\mathrm{3}=\frac{\mid{C}_{\mathrm{1}} −{C}_{\mathrm{2}} \mid}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}\:\:\Rightarrow\mathrm{3}=\frac{\mid\frac{\mathrm{1}}{\mathrm{5}}−{C}_{\mathrm{2}} \mid}{\mathrm{13}/\mathrm{5}} \\ $$$$\:\:\:\:\:\:\Rightarrow\frac{\mathrm{39}}{\mathrm{5}}=\mid\frac{\mathrm{1}}{\mathrm{5}}−{C}_{\mathrm{2}} \mid \\ $$$$\:\:\:\:\:\:\Rightarrow{C}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{39}}{\mathrm{5}}\:\:{or}\:{C}_{\mathrm{2}} =\frac{\mathrm{39}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\Rightarrow{C}_{\mathrm{2}} =−\frac{\mathrm{38}}{\mathrm{5}}\:\:{or}\:\:\:\mathrm{8} \\ $$$${from}\:\left({i}\right)\:{y}=−\frac{\mathrm{12}}{\mathrm{5}}{x}−\frac{\mathrm{34}}{\mathrm{5}}\:\:{and}\:{y}=−\frac{\mathrm{12}}{\mathrm{5}}{x}+\mathrm{8} \\ $$$$\Rightarrow\mathrm{12}{x}+\mathrm{5}{y}+\mathrm{34}=\mathrm{0}\:\:\:\:{and}\:\:\:\mathrm{12}{x}+\mathrm{5}{y}−\mathrm{40}=\mathrm{0} \\ $$
Commented by nadovic last updated on 15/Aug/21
$${Thank}\:{you}\:{Sir} \\ $$