Question Number 151513 by pete last updated on 21/Aug/21
$$\mathrm{Find}\:\mathrm{two}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:{p}\:\mathrm{if}\:\mathrm{the}\:\mathrm{lines} \\ $$$${px}−{y}=\mathrm{0}\:\mathrm{and}\:\mathrm{3}{x}+{y}+\mathrm{1}=\mathrm{0}\:\mathrm{intersect}\:\mathrm{at}\:\mathrm{45}° \\ $$
Answered by Olaf_Thorendsen last updated on 21/Aug/21
$$\Delta_{\mathrm{1}} \::\:{px}−{y}\:=\:\mathrm{0} \\ $$$$\Delta_{\mathrm{2}} \::\:\mathrm{3}{x}+{y}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{1}} } \:=\:\begin{pmatrix}{\mathrm{1}}\\{{p}}\end{pmatrix} \\ $$$$\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{2}} } \:=\:\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{3}}\end{pmatrix} \\ $$$$\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{1}} } \bullet\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{2}} } \:=\:\mid\mid\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{1}} } \mid\mid×\mid\mid\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{2}} } \mid\mid×\mathrm{cos}\left(\widehat {\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{1}} } ,\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{2}} } }\right) \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{1}\right)+{p}\left(−\mathrm{3}\right)\:=\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }×\sqrt{\mathrm{10}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{1}−\mathrm{3}{p}\right) \\ $$$$\Rightarrow\:\mathrm{1}+{p}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{9}{p}^{\mathrm{2}} −\mathrm{6}{p}+\mathrm{1}\right) \\ $$$$\mathrm{2}{p}^{\mathrm{2}} −\mathrm{3}{p}−\mathrm{2}\:=\:\mathrm{0} \\ $$$$\mathrm{2}\left({p}−\mathrm{2}\right)\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$${p}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{or}\:{p}\:=\:\mathrm{2} \\ $$
Commented by pete last updated on 21/Aug/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by nadovic last updated on 21/Aug/21
$${l}_{\mathrm{1}} :\:\:\:\:{px}\:−\:{y}\:=\:\mathrm{0} \\ $$$${gradient},\:{m}_{\mathrm{1}} \:=\:{p} \\ $$$${l}_{\mathrm{2}} :\:\:\:\:\mathrm{3}{x}\:+\:{y}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$${gradient},\:{m}_{\mathrm{2}} \:=\:−\mathrm{3} \\ $$$$\:\Rightarrow\:\:\:\:\:\mid\frac{{m}_{\mathrm{1}} \:−\:{m}_{\mathrm{2}} }{\mathrm{1}\:+\:{m}_{\mathrm{1}} {m}_{\mathrm{2}} }\mid\:\:=\:\:\mathrm{tan}\:\mathrm{45}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{{p}\:−\:\left(−\mathrm{3}\right)}{\mathrm{1}\:+\:{p}\left(−\mathrm{3}\right)}\:\:=\:\pm\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:{p}\:+\:\mathrm{3}\:=\:\pm\mathrm{1}\:\mp\:\mathrm{3}{p}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{p}\:=\:−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{OR}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{p}\:=\:−\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}\:=\:\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\: \\ $$
Commented by bramlexs22 last updated on 21/Aug/21
$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{tan}\:\mathrm{45}°=\mid\frac{\mathrm{m}_{\mathrm{1}} −\mathrm{m}_{\mathrm{2}} }{\mathrm{1}+\mathrm{m}_{\mathrm{1}} \mathrm{m}_{\mathrm{2}} }\mid \\ $$
Commented by nadovic last updated on 21/Aug/21
$${Alright}\:{Sir}.\:{Thank}\:{you}! \\ $$
Commented by pete last updated on 21/Aug/21
$$\mathrm{thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$