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Find-two-possible-values-of-p-if-the-lines-px-y-0-and-3x-y-1-0-intersect-at-45-




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°
Findtwopossiblevaluesofpifthelinespxy=0and3x+y+1=0intersectat45°
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
Δ1:pxy=0Δ2:3x+y+1=0uΔ1=(1p)uΔ2=(13)uΔ1uΔ2=∣∣uΔ1∣∣×∣∣uΔ2∣∣×cos(uΔ1,uΔ2^)(1)(1)+p(3)=1+p2×10×121+p2=15(13p)1+p2=15(9p26p+1)2p23p2=02(p2)(p+12)=0p=12orp=2
Commented by pete last updated on 21/Aug/21
thank you sir
thankyousir
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
l1:pxy=0gradient,m1=pl2:3x+y+1=0gradient,m2=3m1m21+m1m2=tan45°p(3)1+p(3)=±1p+3=±13p4p=2p=12OR2p=4p=2
Commented by bramlexs22 last updated on 21/Aug/21
it should be tan 45°=∣((m_1 −m_2 )/(1+m_1 m_2 ))∣
itshouldbetan45°=∣m1m21+m1m2
Commented by nadovic last updated on 21/Aug/21
Alright Sir. Thank you!
AlrightSir.Thankyou!
Commented by pete last updated on 21/Aug/21
thanks very much sir
thanksverymuchsir

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