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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

(1) mx+2y−1=0 (2) 2x+my+3=0 (3) x−y−3=0 maybe I do not understand what to do, but we have 3 equations in 3 unknown, so we get an unique solution (1) ⇒ y=((1−mx)/2) (2) ⇒ x=((m+6)/(m^2 −4)) ⇒ y=−((3m+2)/(m^2 −4)) (3) ⇒ m=((10)/3) ⇒ x=((21)/(16))∧y=−((27)/(16))

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

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