Question Number 26411 by Tinkutara last updated on 25/Dec/17
$${Show}\:{that}\:{x}^{\mathrm{2}} \:+\:\mathrm{4}{xy}\:−\:\mathrm{2}{y}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:−\:\mathrm{12}{y} \\ $$$$−\:\mathrm{15}\:=\:\mathrm{0}\:{represents}\:{a}\:{pair}\:{of}\:{straight} \\ $$$${lines}\:{and}\:{that}\:{these}\:{lines}\:{together} \\ $$$${with}\:{the}\:{pair}\:{of}\:{lines}\:{x}^{\mathrm{2}} \:+\:\mathrm{4}{xy}\:−\:\mathrm{2}{y}^{\mathrm{2}} \\ $$$$=\:\mathrm{0}\:{form}\:{a}\:{rhombus}. \\ $$
Answered by ajfour last updated on 25/Dec/17
![x^2 +4xy−2y^2 +6x−12y−15=0 determinant ((a,h,g),(h,b,f),(g,f,( c)))= determinant ((1,2,3),(2,(−2),(−6)),(3,(−6),(−15))) =−6+24−18 =0 hence pair of straight lines. If both pair of lines having same two slopes, have a common angular bisector , then they form a rhombus. Here as the other pair passes through origin, we can prove this by showing origin is equidistant from the each line of the first pair. ⇒ (c_1 /( (√(1+m_1 ^2 ))))=(c_2 /( (√(1+m_2 ^2 )))) or c_1 ^2 (1+m_2 ^2 )=c_2 ^2 (1+m_1 ^2 ) ..(i) (y−m_1 x−c_1 )(y−m_2 x−c_2 )=0 c_1 c_2 =((−15)/(−2)) =((15)/2) ; they are roots of z^2 +6z+((15)/2)=0 c_1 ,c_2 =((−6±(√(36−30)))/2)=−3±((√6)/2) m_1 m_2 =(a/b)=−(1/2) , m_1 +m_2 =−((2h)/b)=2 ; ⇒ roots of z^2 −2z−(1/2)=0 m_1 , m_2 =((2±(√(4+2)))/2) =1±((√6)/2) Also m_1 c_2 +m_2 c_1 =((2g)/b) =(6/(−2))=−3 (1+((√6)/2))(−3−((√6)/2))+(1−((√6)/2))(−3+((√6)/2)) =−3−((√6)/2)−((3(√6))/2)−(3/2)−3+((√6)/2)+((3(√6))/2)−(3/2) =−9 , so (1+((√6)/2))(−3+((√6)/2))+(1−((√6)/2))(−3−((√6)/2)) =−(3/2)−(√6) −(3/2)+(√6) =−3 ....(ii) hence if m_1 =1−((√6)/2) , c_2 =−3−((√6)/2) eq. of angular bisector of first pair of lines are thus ((y−m_1 x−c_1 )/( (√(1+m_1 ^2 ))))=±((y−m_2 x−c_2 )/( (√(1+m_2 ^2 )))) as to form a rhombus one bisector line must pass through origin , so c_1 (√(1+m_2 ^2 )) =c_2 (√(1+m_1 ^2 )) as c_1 c_2 > 0 ⇒ c_1 ^2 −c_2 ^2 =m_1 ^2 c_2 ^2 −m_2 ^2 c_1 ^2 (c_1 +c_2 )(c_1 −c_2 )=(m_1 c_2 +m_2 c_1 )× (m_1 c_2 −m_2 c_1 ) l.h.s. =(−6)((√6)) =−6(√6) r.h.s.=(−3)(2(√6))=−6(√6) [see (ii)] hence the lines form a rhombus.](https://www.tinkutara.com/question/Q26426.png)
$${x}^{\mathrm{2}} +\mathrm{4}{xy}−\mathrm{2}{y}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{12}{y}−\mathrm{15}=\mathrm{0} \\ $$$$\begin{vmatrix}{{a}}&{{h}}&{{g}}\\{{h}}&{{b}}&{{f}}\\{{g}}&{{f}}&{\:{c}}\end{vmatrix}=\begin{vmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{2}}&{−\mathrm{2}}&{−\mathrm{6}}\\{\mathrm{3}}&{−\mathrm{6}}&{−\mathrm{15}}\end{vmatrix} \\ $$$$=−\mathrm{6}+\mathrm{24}−\mathrm{18}\:=\mathrm{0} \\ $$$${hence}\:{pair}\:{of}\:{straight}\:{lines}. \\ $$$$ \\ $$$${If}\:{both}\:{pair}\:{of}\:{lines}\:{having} \\ $$$${same}\:{two}\:{slopes},\:{have}\:{a}\:{common} \\ $$$${angular}\:{bisector}\:,\:{then}\:{they}\:{form} \\ $$$${a}\:{rhombus}.\:{Here}\:{as}\:{the}\:{other} \\ $$$${pair}\:{passes}\:{through}\:{origin},\:{we} \\ $$$${can}\:{prove}\:{this}\:{by}\:{showing}\:{origin} \\ $$$${is}\:{equidistant}\:{from}\:{the}\:{each} \\ $$$${line}\:{of}\:{the}\:{first}\:{pair}. \\ $$$$\Rightarrow\:\frac{{c}_{\mathrm{1}} }{\:\sqrt{\mathrm{1}+{m}_{\mathrm{1}} ^{\mathrm{2}} }}=\frac{{c}_{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{m}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$${or}\:\:\:{c}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{1}+{m}_{\mathrm{2}} ^{\mathrm{2}} \right)={c}_{\mathrm{2}} ^{\mathrm{2}} \left(\mathrm{1}+{m}_{\mathrm{1}} ^{\mathrm{2}} \right)\:\:..\left({i}\right) \\ $$$$\:\:\:\left({y}−{m}_{\mathrm{1}} {x}−{c}_{\mathrm{1}} \right)\left({y}−{m}_{\mathrm{2}} {x}−{c}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$${c}_{\mathrm{1}} {c}_{\mathrm{2}} =\frac{−\mathrm{15}}{−\mathrm{2}}\:=\frac{\mathrm{15}}{\mathrm{2}}\:\:;\:{they}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} +\mathrm{6}{z}+\frac{\mathrm{15}}{\mathrm{2}}=\mathrm{0} \\ $$$${c}_{\mathrm{1}} ,{c}_{\mathrm{2}} =\frac{−\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{30}}}{\mathrm{2}}=−\mathrm{3}\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${m}_{\mathrm{1}} {m}_{\mathrm{2}} =\frac{{a}}{{b}}=−\frac{\mathrm{1}}{\mathrm{2}}\:,\: \\ $$$${m}_{\mathrm{1}} +{m}_{\mathrm{2}} =−\frac{\mathrm{2}{h}}{{b}}=\mathrm{2}\:;\:\Rightarrow\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\:\: \\ $$$${m}_{\mathrm{1}} ,\:{m}_{\mathrm{2}} =\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{2}}}{\mathrm{2}}\:=\mathrm{1}\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${Also}\:{m}_{\mathrm{1}} {c}_{\mathrm{2}} +{m}_{\mathrm{2}} {c}_{\mathrm{1}} =\frac{\mathrm{2}{g}}{{b}}\:=\frac{\mathrm{6}}{−\mathrm{2}}=−\mathrm{3} \\ $$$$\left(\mathrm{1}+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)\left(−\mathrm{3}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)+\left(\mathrm{1}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)\left(−\mathrm{3}+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right) \\ $$$$=−\mathrm{3}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{3}+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=−\mathrm{9}\:\:,\:{so} \\ $$$$\left(\mathrm{1}+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)\left(−\mathrm{3}+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)+\left(\mathrm{1}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)\left(−\mathrm{3}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\mathrm{6}}\:\:−\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\mathrm{6}}\:=−\mathrm{3}\:\:\:….\left({ii}\right) \\ $$$${hence}\:{if}\:{m}_{\mathrm{1}} =\mathrm{1}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:,\:{c}_{\mathrm{2}} =−\mathrm{3}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${eq}.\:{of}\:{angular}\:{bisector}\:{of}\:{first} \\ $$$${pair}\:{of}\:{lines}\:{are}\:{thus} \\ $$$$\frac{{y}−{m}_{\mathrm{1}} {x}−{c}_{\mathrm{1}} }{\:\sqrt{\mathrm{1}+{m}_{\mathrm{1}} ^{\mathrm{2}} }}=\pm\frac{{y}−{m}_{\mathrm{2}} {x}−{c}_{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{m}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$${as}\:{to}\:{form}\:{a}\:{rhombus}\:{one}\: \\ $$$${bisector}\:{line}\:{must}\:{pass}\:{through} \\ $$$${origin}\:,\:{so} \\ $$$${c}_{\mathrm{1}} \sqrt{\mathrm{1}+{m}_{\mathrm{2}} ^{\mathrm{2}} }\:={c}_{\mathrm{2}} \sqrt{\mathrm{1}+{m}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$${as}\:\:{c}_{\mathrm{1}} {c}_{\mathrm{2}} \:>\:\mathrm{0} \\ $$$$\Rightarrow\:{c}_{\mathrm{1}} ^{\mathrm{2}} −{c}_{\mathrm{2}} ^{\mathrm{2}} ={m}_{\mathrm{1}} ^{\mathrm{2}} {c}_{\mathrm{2}} ^{\mathrm{2}} −{m}_{\mathrm{2}} ^{\mathrm{2}} {c}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} \right)\left({c}_{\mathrm{1}} −{c}_{\mathrm{2}} \right)=\left({m}_{\mathrm{1}} {c}_{\mathrm{2}} +{m}_{\mathrm{2}} {c}_{\mathrm{1}} \right)× \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({m}_{\mathrm{1}} {c}_{\mathrm{2}} −{m}_{\mathrm{2}} {c}_{\mathrm{1}} \right) \\ $$$${l}.{h}.{s}.\:=\left(−\mathrm{6}\right)\left(\sqrt{\mathrm{6}}\right)\:=−\mathrm{6}\sqrt{\mathrm{6}} \\ $$$${r}.{h}.{s}.=\left(−\mathrm{3}\right)\left(\mathrm{2}\sqrt{\mathrm{6}}\right)=−\mathrm{6}\sqrt{\mathrm{6}}\:\:\:\left[{see}\:\left({ii}\right)\right] \\ $$$${hence}\:{the}\:{lines}\:{form}\:{a}\:{rhombus}. \\ $$
Commented by Tinkutara last updated on 25/Dec/17
Thank you Sir!