Question Number 212179 by cherokeesay last updated on 05/Oct/24
Answered by cemoosky last updated on 05/Oct/24
$${x}\:=\:\mathrm{6}.\mathrm{6} \\ $$
Answered by mr W last updated on 06/Oct/24
Commented by mr W last updated on 06/Oct/24
$${AD}:\:{y}={mx} \\ $$$${y}=\mathrm{1}−{x} \\ $$$$\mathrm{1}−{x}_{{B}} ={mx}_{{B}} \:\Rightarrow{x}_{{B}} =\frac{\mathrm{1}}{\mathrm{1}+{m}} \\ $$$${y}=−\mathrm{1}+{x} \\ $$$$−\mathrm{1}+{x}_{{C}} ={mx}_{{C}} \:\Rightarrow{x}_{{C}} =\frac{\mathrm{1}}{\mathrm{1}−{m}} \\ $$$$\frac{{a}}{{b}}=\frac{{x}_{{B}} }{{x}_{{C}} −{x}_{{B}} }=\frac{\frac{\mathrm{1}}{\mathrm{1}+{m}}}{\frac{\mathrm{1}}{\mathrm{1}−{m}}−\frac{\mathrm{1}}{\mathrm{1}+{m}}}=\frac{\mathrm{1}−{m}}{\mathrm{2}{m}} \\ $$$$\Rightarrow{m}=\frac{{b}}{\mathrm{2}{a}+{b}} \\ $$$$\frac{{x}}{{a}}=\frac{\mathrm{2}−{x}_{{C}} }{{x}_{{B}} }=\frac{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{1}−{m}}}{\frac{\mathrm{1}}{\mathrm{1}+{m}}}=\frac{\left(\mathrm{1}−\mathrm{2}{m}\right)\left(\mathrm{1}+{m}\right)}{\mathrm{1}−{m}} \\ $$$${x}=\frac{\left(\mathrm{1}−\mathrm{2}{m}\right)\left(\mathrm{1}+{m}\right){b}}{\mathrm{2}{m}} \\ $$$$\Rightarrow{x}=\frac{\left({a}+{b}\right)\left(\mathrm{2}{a}−{b}\right)}{\mathrm{2}{a}+{b}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{6}+\mathrm{4}\right)\left(\mathrm{2}×\mathrm{6}−\mathrm{4}\right)}{\mathrm{2}×\mathrm{6}+\mathrm{4}}=\mathrm{5} \\ $$
Commented by cherokeesay last updated on 06/Oct/24
$${thank}\:{you}\:{master}\:! \\ $$