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The-lines-ax-2y-1-0-bx-3y-1-0-and-cx-4y-1-0-are-concurrent-if-a-b-c-are-in-G-P-




Question Number 76245 by vishalbhardwaj last updated on 25/Dec/19
The lines ax+2y+1=0, bx+3y+1=0  and cx+4y+1=0 are concurrent  if a, b, c are in G.P. ??
$$\mathrm{The}\:\mathrm{lines}\:{ax}+\mathrm{2}{y}+\mathrm{1}=\mathrm{0},\:{bx}+\mathrm{3}{y}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{and}\:{cx}+\mathrm{4}{y}+\mathrm{1}=\mathrm{0}\:\mathrm{are}\:\mathrm{concurrent} \\ $$$$\mathrm{if}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{in}\:\mathrm{G}.\mathrm{P}.\:?? \\ $$
Answered by MJS last updated on 25/Dec/19
no, they are parallel if a=2k∧b=3k∧c=4k  2kx+2y+1=0 ⇔ kx+y+(1/2)=0  3kx+3y+1=0 ⇔ kx+y+(1/3)=0  4kx+4y+1=0 ⇔ kx+y+(1/4)=0  since the constant factors are never the  same the lines are never concurrent
$$\mathrm{no},\:\mathrm{they}\:\mathrm{are}\:\mathrm{parallel}\:\mathrm{if}\:{a}=\mathrm{2}{k}\wedge{b}=\mathrm{3}{k}\wedge{c}=\mathrm{4}{k} \\ $$$$\mathrm{2}{kx}+\mathrm{2}{y}+\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:{kx}+{y}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{3}{kx}+\mathrm{3}{y}+\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:{kx}+{y}+\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$$\mathrm{4}{kx}+\mathrm{4}{y}+\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:{kx}+{y}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{since}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{factors}\:\mathrm{are}\:\mathrm{never}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{the}\:\mathrm{lines}\:\mathrm{are}\:\mathrm{never}\:\mathrm{concurrent} \\ $$

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