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