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if-the-angle-between-kx-5y-1-kx-2y-2-equal-60-then-k-help-me-sir-

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Question Number 112908 by mohammad17 last updated on 10/Sep/20
if the angle between( kx+5y=1 , kx−2y=2)equal 60^° then k=? help me sir
$${if}\:{the}\:{angle}\:{between}\left(\:{kx}+\mathrm{5}{y}=\mathrm{1}\:,\:{kx}−\mathrm{2}{y}=\mathrm{2}\right){equal}\:\mathrm{60}^{°} {then}\:{k}=? \\ $$$$ \\ $$$${help}\:{me}\:{sir} \\ $$
Commented by mohammad17 last updated on 10/Sep/20
please sir im very need this
$${please}\:{sir}\:{im}\:{very}\:{need}\:{this} \\ $$
Answered by ajfour last updated on 10/Sep/20
y=mx+c m_1 =−(k/5) , m_2 =(k/2) (((k/2)−(−(k/5)))/(1+((k/2))(−(k/5))))=tan 60° = (√3) ⇒ 7k=(√3)(10−k^2 ) ⇒ k^2 +((7k)/( (√3)))−10 = 0 k=−(7/(2(√3)))±(√(((49)/(12))+10)) k=−((10(√3))/3) or = (√3) .
$${y}={mx}+{c} \\ $$$${m}_{\mathrm{1}} =−\frac{{k}}{\mathrm{5}}\:\:,\:\:{m}_{\mathrm{2}} =\frac{{k}}{\mathrm{2}} \\ $$$$\frac{\frac{{k}}{\mathrm{2}}−\left(−\frac{{k}}{\mathrm{5}}\right)}{\mathrm{1}+\left(\frac{{k}}{\mathrm{2}}\right)\left(−\frac{{k}}{\mathrm{5}}\right)}=\mathrm{tan}\:\mathrm{60}°\:=\:\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:\mathrm{7}{k}=\sqrt{\mathrm{3}}\left(\mathrm{10}−{k}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\:{k}^{\mathrm{2}} +\frac{\mathrm{7}{k}}{\:\sqrt{\mathrm{3}}}−\mathrm{10}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{k}=−\frac{\mathrm{7}}{\mathrm{2}\sqrt{\mathrm{3}}}\pm\sqrt{\frac{\mathrm{49}}{\mathrm{12}}+\mathrm{10}} \\ $$$$\:\:\:\:\:\:{k}=−\frac{\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\:{or}\:\:=\:\sqrt{\mathrm{3}}\:. \\ $$
Answered by mathmax by abdo last updated on 11/Sep/20
u(−5,k) and v(2,k) (u,v) =(π/3) tan(u,v) =((det(u,v))/(u.v)) =( determinant (((−5 2)),((k k)))/(k^2 −10)) =(√(3 )) ⇒((−7k)/(k^2 −10)) =(√3) ⇒ −7k =(√3)k^2 −10(√3) ⇒(√3)k^2 +7k −10(√3)=0 Δ =49+4.10.3 =49 +120 =169 ⇒k_1 =((−7+13)/(2(√3))) =(6/(2(√3))) =(√3) k_2 =((−7−13)/(2(√3))) =((−20)/(2(√3))) =((−10)/( (√3)))
$$\mathrm{u}\left(−\mathrm{5},\mathrm{k}\right)\:\:\mathrm{and}\:\mathrm{v}\left(\mathrm{2},\mathrm{k}\right)\:\:\:\left(\mathrm{u},\mathrm{v}\right)\:=\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{tan}\left(\mathrm{u},\mathrm{v}\right)\:=\frac{\mathrm{det}\left(\mathrm{u},\mathrm{v}\right)}{\mathrm{u}.\mathrm{v}}\:=\frac{\begin{vmatrix}{−\mathrm{5}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{k}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{k}}\end{vmatrix}}{\mathrm{k}^{\mathrm{2}} −\mathrm{10}}\:=\sqrt{\mathrm{3}\:}\:\Rightarrow\frac{−\mathrm{7k}}{\mathrm{k}^{\mathrm{2}} −\mathrm{10}}\:=\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$$−\mathrm{7k}\:=\sqrt{\mathrm{3}}\mathrm{k}^{\mathrm{2}} −\mathrm{10}\sqrt{\mathrm{3}}\:\Rightarrow\sqrt{\mathrm{3}}\mathrm{k}^{\mathrm{2}} \:+\mathrm{7k}\:−\mathrm{10}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\Delta\:=\mathrm{49}+\mathrm{4}.\mathrm{10}.\mathrm{3}\:=\mathrm{49}\:+\mathrm{120}\:=\mathrm{169}\:\Rightarrow\mathrm{k}_{\mathrm{1}} =\frac{−\mathrm{7}+\mathrm{13}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\frac{\mathrm{6}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\sqrt{\mathrm{3}} \\ $$$$\mathrm{k}_{\mathrm{2}} =\frac{−\mathrm{7}−\mathrm{13}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\frac{−\mathrm{20}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\frac{−\mathrm{10}}{\:\sqrt{\mathrm{3}}} \\ $$

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