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Question Number 63522 by Tawa1 last updated on 05/Jul/19

Commented by Tawa1 last updated on 05/Jul/19

Find the real parameter “m” such that cross cutting of mx + 2y - 1 = 0 and 2x + my + 3 = 0 give slopes equation belongs x - y - 3 = 0

Commented by MJS last updated on 05/Jul/19

$$\left(1\right)mx+2y-1=0$$$$\left(2\right)2x+my+3=0$$$$\left(3\right)x-y-3=0$$$$\mathrm{maybe}\mathrm{I}\mathrm{do}\mathrm{not}\mathrm{understand}\mathrm{what}\mathrm{to}\mathrm{do},\mathrm{but}$$$$\mathrm{we}\mathrm{have}3\mathrm{equations}\mathrm{in}3\mathrm{unknown},\mathrm{so}\mathrm{we}\mathrm{get}$$$$\mathrm{an}\mathrm{unique}\mathrm{solution}$$$$\left(1\right)\Rightarrow y=\frac{1-mx}{2}$$$$\left(2\right)\Rightarrow x=\frac{m+6}{m^2-4}\Rightarrow y=-\frac{3m+2}{m^2-4}$$$$\left(3\right)\Rightarrow m=\frac{10}{3}\Rightarrow x=\frac{21}{16}\wedge y=-\frac{27}{16}$$

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