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Question Number 54322 by peter frank last updated on 02/Feb/19

Answered by kaivan.ahmadi last updated on 02/Feb/19

ax+by=c and bx−ay=d are perpendicular abx^2 +(b^2 −a^2 )xy−aby^2 =cd ab=6 k=−ab=−6

$$\mathrm{ax}+\mathrm{by}=\mathrm{c}\:\mathrm{and}\:\mathrm{bx}−\mathrm{ay}=\mathrm{d}\:\mathrm{are}\:\mathrm{perpendicular} \\ $$$$\mathrm{abx}^{\mathrm{2}} +\left(\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)\mathrm{xy}−\mathrm{aby}^{\mathrm{2}} =\mathrm{cd} \\ $$$$\mathrm{ab}=\mathrm{6} \\ $$$$\mathrm{k}=−\mathrm{ab}=−\mathrm{6} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Feb/19

ky^2 −xy+6x^2 =0 y^2 +(−(1/k))xy+((6/k))x^2 =0 (y−m_1 x)(y−m_2 x)=0 y^2 −(m_1 +m_2 )xy+m_1 m_2 x^2 =0 −(m_1 +m_2 )xy=−(1/k)xy m_1 +m_2 =(1/k) m_1 m_2 =(6/k) i)two⊥lines m_1 m_2 =(6/k)=−1 so k=−6 iii)tan(π/4)=((m_1 −m_2 )/(1+m_1 m_2 ))=((√((m_1 +m_2 )^2 −4m_1 m_2 ))/(1+m_1 m_2 )) 1=((√((1/k^2 )−((24)/k)))/(1+(6/k))) 1+((12)/k)+((36)/k^2 )=(1/k^2 )−((24)/k) k^2 +12k+36=1−24k k^2 +36k+35=0 (k+1)(k+35)=0 k=−1 and k=−35

$${ky}^{\mathrm{2}} −{xy}+\mathrm{6}{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}^{\mathrm{2}} +\left(−\frac{\mathrm{1}}{{k}}\right){xy}+\left(\frac{\mathrm{6}}{{k}}\right){x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({y}−{m}_{\mathrm{1}} {x}\right)\left({y}−{m}_{\mathrm{2}} {x}\right)=\mathrm{0} \\ $$$${y}^{\mathrm{2}} −\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){xy}+{m}_{\mathrm{1}} {m}_{\mathrm{2}} {x}^{\mathrm{2}} =\mathrm{0} \\ $$$$−\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){xy}=−\frac{\mathrm{1}}{{k}}{xy} \\ $$$${m}_{\mathrm{1}} +{m}_{\mathrm{2}} =\frac{\mathrm{1}}{{k}} \\ $$$${m}_{\mathrm{1}} {m}_{\mathrm{2}} =\frac{\mathrm{6}}{{k}} \\ $$$$\left.{i}\right){two}\bot{lines}\:\:{m}_{\mathrm{1}} {m}_{\mathrm{2}} =\frac{\mathrm{6}}{{k}}=−\mathrm{1}\:\:\:{so}\:{k}=−\mathrm{6} \\ $$$$\left.{iii}\right){tan}\frac{\pi}{\mathrm{4}}=\frac{{m}_{\mathrm{1}} −{m}_{\mathrm{2}} }{\mathrm{1}+{m}_{\mathrm{1}} {m}_{\mathrm{2}} }=\frac{\sqrt{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{m}_{\mathrm{1}} {m}_{\mathrm{2}} }}{\mathrm{1}+{m}_{\mathrm{1}} {m}_{\mathrm{2}} } \\ $$$$\mathrm{1}=\frac{\sqrt{\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\frac{\mathrm{24}}{{k}}}}{\mathrm{1}+\frac{\mathrm{6}}{{k}}} \\ $$$$\mathrm{1}+\frac{\mathrm{12}}{{k}}+\frac{\mathrm{36}}{{k}^{\mathrm{2}} }=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\frac{\mathrm{24}}{{k}} \\ $$$${k}^{\mathrm{2}} +\mathrm{12}{k}+\mathrm{36}=\mathrm{1}−\mathrm{24}{k} \\ $$$${k}^{\mathrm{2}} +\mathrm{36}{k}+\mathrm{35}=\mathrm{0} \\ $$$$\left({k}+\mathrm{1}\right)\left({k}+\mathrm{35}\right)=\mathrm{0} \\ $$$${k}=−\mathrm{1}\:{and}\:{k}=−\mathrm{35} \\ $$

Commented by peter frank last updated on 02/Feb/19

thank you very much

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

Answered by mr W last updated on 02/Feb/19

6x^2 −xy+ky^2 =(ax+by)(cx+dy) 6x^2 −xy+ky^2 =acx^2 +(ad+bc)xy+bdy^2 ac=6 ad+bc=−1 bd=k set a=1 ⇒c=6 d+6b=−1 bd=k b(6b+1)+k=0 6b^2 +b+k=0 ⇒b=((−1±(√(1−24k)))/(12)) (i): (a/b)=−(d/c)⇒ac=−bd ⇒k=bd=−ac=−6 (ii): D=1−24k≥0 ⇒k≤(1/(24)) i.e. if k>(1/(24))⇒no graph if k=(1/(24))⇒the graph is a single line if k<(1/(24))⇒the graph is two lines (iii): m_1 =(a/b)=tan θ_1 m_2 =(c/d)=tan θ_2 θ_1 −θ_2 =±(π/4) tan (θ_1 −θ_2 )=±1 ((m_1 −m_2 )/(1+m_1 m_2 ))=±1 (((a/b)−(c/d))/(1+((ac)/(bd))))=±1 ((ad−bc)/(ac+bd))=±1 ((1+2bc)/(6+k))=±1 (√(1−24k))=±(6+k) 1−24k=36+12k+k^2 35+36k+k^2 =0 k=((−36±(√(36^2 −4×35)))/2)=−18±17 ⇒k=−1 or −35

$$\mathrm{6}{x}^{\mathrm{2}} −{xy}+{ky}^{\mathrm{2}} =\left({ax}+{by}\right)\left({cx}+{dy}\right) \\ $$$$\mathrm{6}{x}^{\mathrm{2}} −{xy}+{ky}^{\mathrm{2}} ={acx}^{\mathrm{2}} +\left({ad}+{bc}\right){xy}+{bdy}^{\mathrm{2}} \\ $$$${ac}=\mathrm{6} \\ $$$${ad}+{bc}=−\mathrm{1} \\ $$$${bd}={k} \\ $$$$ \\ $$$${set}\:{a}=\mathrm{1} \\ $$$$\Rightarrow{c}=\mathrm{6} \\ $$$${d}+\mathrm{6}{b}=−\mathrm{1} \\ $$$${bd}={k} \\ $$$${b}\left(\mathrm{6}{b}+\mathrm{1}\right)+{k}=\mathrm{0} \\ $$$$\mathrm{6}{b}^{\mathrm{2}} +{b}+{k}=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{24}{k}}}{\mathrm{12}} \\ $$$$ \\ $$$$\left({i}\right): \\ $$$$\frac{{a}}{{b}}=−\frac{{d}}{{c}}\Rightarrow{ac}=−{bd} \\ $$$$\Rightarrow{k}={bd}=−{ac}=−\mathrm{6} \\ $$$$ \\ $$$$\left({ii}\right): \\ $$$${D}=\mathrm{1}−\mathrm{24}{k}\geqslant\mathrm{0} \\ $$$$\Rightarrow{k}\leqslant\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${i}.{e}.\:{if}\:{k}>\frac{\mathrm{1}}{\mathrm{24}}\Rightarrow{no}\:{graph} \\ $$$${if}\:{k}=\frac{\mathrm{1}}{\mathrm{24}}\Rightarrow{the}\:{graph}\:{is}\:{a}\:{single}\:{line} \\ $$$${if}\:{k}<\frac{\mathrm{1}}{\mathrm{24}}\Rightarrow{the}\:{graph}\:{is}\:{two}\:{lines} \\ $$$$ \\ $$$$\left({iii}\right): \\ $$$${m}_{\mathrm{1}} =\frac{{a}}{{b}}=\mathrm{tan}\:\theta_{\mathrm{1}} \\ $$$${m}_{\mathrm{2}} =\frac{{c}}{{d}}=\mathrm{tan}\:\theta_{\mathrm{2}} \\ $$$$\theta_{\mathrm{1}} −\theta_{\mathrm{2}} =\pm\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{tan}\:\left(\theta_{\mathrm{1}} −\theta_{\mathrm{2}} \right)=\pm\mathrm{1} \\ $$$$\frac{{m}_{\mathrm{1}} −{m}_{\mathrm{2}} }{\mathrm{1}+{m}_{\mathrm{1}} {m}_{\mathrm{2}} }=\pm\mathrm{1} \\ $$$$\frac{\frac{{a}}{{b}}−\frac{{c}}{{d}}}{\mathrm{1}+\frac{{ac}}{{bd}}}=\pm\mathrm{1} \\ $$$$\frac{{ad}−{bc}}{{ac}+{bd}}=\pm\mathrm{1} \\ $$$$\frac{\mathrm{1}+\mathrm{2}{bc}}{\mathrm{6}+{k}}=\pm\mathrm{1} \\ $$$$\sqrt{\mathrm{1}−\mathrm{24}{k}}=\pm\left(\mathrm{6}+{k}\right) \\ $$$$\mathrm{1}−\mathrm{24}{k}=\mathrm{36}+\mathrm{12}{k}+{k}^{\mathrm{2}} \\ $$$$\mathrm{35}+\mathrm{36}{k}+{k}^{\mathrm{2}} =\mathrm{0} \\ $$$${k}=\frac{−\mathrm{36}\pm\sqrt{\mathrm{36}^{\mathrm{2}} −\mathrm{4}×\mathrm{35}}}{\mathrm{2}}=−\mathrm{18}\pm\mathrm{17} \\ $$$$\Rightarrow{k}=−\mathrm{1}\:{or}\:−\mathrm{35} \\ $$

Commented by mr W last updated on 02/Feb/19

Commented by mr W last updated on 02/Feb/19

Commented by mr W last updated on 02/Feb/19

Commented by mr W last updated on 02/Feb/19

Commented by peter frank last updated on 02/Feb/19

thank you very much sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}. \\ $$

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