Question-63522

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) m x + 2 y - 1 = 0

(2) 2 x + m y + 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) \Rightarrow y = \frac{1 - m x}{2}

(2) \Rightarrow x = \frac{m + 6}{m^{2} - 4} \Rightarrow y = - \frac{3 m + 2}{m^{2} - 4}

(3) \Rightarrow m = \frac{10}{3} \Rightarrow x = \frac{21}{16} \land y = - \frac{27}{16}

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