Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 151513 by pete last updated on 21/Aug/21

Find two possible values of p if the lines px−y=0 and 3x+y+1=0 intersect at 45°

$$\mathrm{Find}\:\mathrm{two}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:{p}\:\mathrm{if}\:\mathrm{the}\:\mathrm{lines} \\ $$$${px}−{y}=\mathrm{0}\:\mathrm{and}\:\mathrm{3}{x}+{y}+\mathrm{1}=\mathrm{0}\:\mathrm{intersect}\:\mathrm{at}\:\mathrm{45}° \\ $$

Answered by Olaf_Thorendsen last updated on 21/Aug/21

Δ_1 : px−y = 0 Δ_2 : 3x+y+1 = 0 u_Δ_1 ^→ = ((1),(p) ) u_Δ_2 ^→ = ((1),((−3)) ) u_Δ_1 ^→ •u_Δ_2 ^→ = ∣∣u_Δ_1 ^→ ∣∣×∣∣u_Δ_2 ^→ ∣∣×cos(u_Δ_1 ^→ ,u_Δ_2 ^→ ^(�) ) (1)(1)+p(−3) = (√(1+p^2 ))×(√(10))×(1/( (√2))) (√(1+p^2 )) = (1/( (√5)))(1−3p) ⇒ 1+p^2 = (1/5)(9p^2 −6p+1) 2p^2 −3p−2 = 0 2(p−2)(p+(1/2)) = 0 p = −(1/2) or p = 2

$$\Delta_{\mathrm{1}} \::\:{px}−{y}\:=\:\mathrm{0} \\ $$$$\Delta_{\mathrm{2}} \::\:\mathrm{3}{x}+{y}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{1}} } \:=\:\begin{pmatrix}{\mathrm{1}}\\{{p}}\end{pmatrix} \\ $$$$\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{2}} } \:=\:\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{3}}\end{pmatrix} \\ $$$$\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{1}} } \bullet\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{2}} } \:=\:\mid\mid\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{1}} } \mid\mid×\mid\mid\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{2}} } \mid\mid×\mathrm{cos}\left(\widehat {\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{1}} } ,\overset{\rightarrow} {{u}}_{\Delta_{\mathrm{2}} } }\right) \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{1}\right)+{p}\left(−\mathrm{3}\right)\:=\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }×\sqrt{\mathrm{10}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{1}−\mathrm{3}{p}\right) \\ $$$$\Rightarrow\:\mathrm{1}+{p}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{9}{p}^{\mathrm{2}} −\mathrm{6}{p}+\mathrm{1}\right) \\ $$$$\mathrm{2}{p}^{\mathrm{2}} −\mathrm{3}{p}−\mathrm{2}\:=\:\mathrm{0} \\ $$$$\mathrm{2}\left({p}−\mathrm{2}\right)\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$${p}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{or}\:{p}\:=\:\mathrm{2} \\ $$

Commented by pete last updated on 21/Aug/21

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by nadovic last updated on 21/Aug/21

l_1 : px − y = 0 gradient, m_1 = p l_2 : 3x + y + 1 = 0 gradient, m_2 = −3 ⇒ ∣((m_1 − m_2 )/(1 + m_1 m_2 ))∣ = tan 45° ((p − (−3))/(1 + p(−3))) = ±1 p + 3 = ±1 ∓ 3p 4p = −2 p = −(1/2) OR −2p = −4 p = 2

$${l}_{\mathrm{1}} :\:\:\:\:{px}\:−\:{y}\:=\:\mathrm{0} \\ $$$${gradient},\:{m}_{\mathrm{1}} \:=\:{p} \\ $$$${l}_{\mathrm{2}} :\:\:\:\:\mathrm{3}{x}\:+\:{y}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$${gradient},\:{m}_{\mathrm{2}} \:=\:−\mathrm{3} \\ $$$$\:\Rightarrow\:\:\:\:\:\mid\frac{{m}_{\mathrm{1}} \:−\:{m}_{\mathrm{2}} }{\mathrm{1}\:+\:{m}_{\mathrm{1}} {m}_{\mathrm{2}} }\mid\:\:=\:\:\mathrm{tan}\:\mathrm{45}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{{p}\:−\:\left(−\mathrm{3}\right)}{\mathrm{1}\:+\:{p}\left(−\mathrm{3}\right)}\:\:=\:\pm\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:{p}\:+\:\mathrm{3}\:=\:\pm\mathrm{1}\:\mp\:\mathrm{3}{p}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{p}\:=\:−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{OR}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{p}\:=\:−\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}\:=\:\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Commented by bramlexs22 last updated on 21/Aug/21

it should be tan 45°=∣((m_1 −m_2 )/(1+m_1 m_2 ))∣

$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{tan}\:\mathrm{45}°=\mid\frac{\mathrm{m}_{\mathrm{1}} −\mathrm{m}_{\mathrm{2}} }{\mathrm{1}+\mathrm{m}_{\mathrm{1}} \mathrm{m}_{\mathrm{2}} }\mid \\ $$

Commented by nadovic last updated on 21/Aug/21

Alright Sir. Thank you!

$${Alright}\:{Sir}.\:{Thank}\:{you}! \\ $$

Commented by pete last updated on 21/Aug/21

thanks very much sir

$$\mathrm{thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com