Question Number 130263 by pete last updated on 23/Jan/21
Answered by Olaf last updated on 23/Jan/21
$$\mathrm{B}\begin{pmatrix}{{x}_{\mathrm{B}} \:=\:\mathrm{12}−\mathrm{2}{y}_{\mathrm{B}} \:=\:\mathrm{12}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{E}\begin{pmatrix}{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{{y}_{\mathrm{E}} }{{x}_{\mathrm{B}} }\:\Rightarrow\:{y}_{\mathrm{E}} \:=\:\mathrm{6}}\end{pmatrix} \\ $$$$\mathrm{A}\begin{pmatrix}{{x}_{\mathrm{E}} \:=\:\frac{{x}_{\mathrm{A}} +{x}_{\mathrm{B}} }{\mathrm{2}}\:\Rightarrow\:{x}_{\mathrm{A}} \:=\:−\mathrm{12}}\\{{y}_{\mathrm{A}} \:=\:\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}}{x}_{\mathrm{A}} \:=\:\mathrm{12}\:}\end{pmatrix} \\ $$$$\left(\mathrm{M}\right)\::\:{y}\:=\:−\frac{\mathrm{1}}{−\frac{\mathrm{1}}{\mathrm{2}}}{x}+{b}\:=\:\mathrm{2}{x}+{b} \\ $$$$\left(\mathrm{M}\right)\::\:{y}\:=\:\mathrm{2}{x}+\mathrm{36} \\ $$
Commented by pete last updated on 23/Jan/21
$$\mathrm{thanks}\:\mathrm{sir},\:\mathrm{but}\:\mathrm{please}\:\mathrm{do}\:\mathrm{give}\:\mathrm{me}\:\mathrm{some} \\ $$$$\mathrm{further}\:\mathrm{explanations}\:\:\mathrm{from}\:\mathrm{the}\:\mathrm{third}\:\mathrm{bracket} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{end}. \\ $$