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Question Number 130263 by pete last updated on 23/Jan/21

Answered by Olaf last updated on 23/Jan/21

$$\mathrm{B}\left(\begin{array}{c}{x}_{\mathrm{B}}=12-2y_{\mathrm{B}}=12\\ 0\end{array}\right)$$$$\mathrm{E}\left(\begin{array}{c}0\\ \frac{1}{2}=\frac{y_{\mathrm{E}}}{x_{\mathrm{B}}}\Rightarrow y_{\mathrm{E}}=6\end{array}\right)$$$$\mathrm{A}\left(\begin{array}{c}{x}_{\mathrm{E}}=\frac{x_{\mathrm{A}}+x_{\mathrm{B}}}{2}\Rightarrow x_{\mathrm{A}}=-12\\ y_{\mathrm{A}}=6-\frac{1}{2}{x}_{\mathrm{A}}=12\end{array}\right)$$$$\left(\mathrm{M}\right):y=-\frac{1}{-\frac{1}{2}}x+b=2x+b$$$$\left(\mathrm{M}\right):y=2x+36$$

Commented by pete last updated on 23/Jan/21

$$\mathrm{thanks}\mathrm{sir},\mathrm{but}\mathrm{please}\mathrm{do}\mathrm{give}\mathrm{me}\mathrm{some}$$$$\mathrm{further}\mathrm{explanations}\mathrm{from}\mathrm{the}\mathrm{third}\mathrm{bracket}$$$$\mathrm{to}\mathrm{the}\mathrm{end}.$$

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