Question Number 38878 by Rio Mike last updated on 30/Jun/18
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{on}\:\mathrm{a}\:\mathrm{line}\:\mathrm{joining} \\ $$$$\mathrm{the}\:\mathrm{points}\:\mathrm{A}\left(\mathrm{2}{x},\mathrm{4}\right),{B}\left({x},\mathrm{3}\right)\:{and}\: \\ $$$${C}\left(\mathrm{4},\mathrm{3}\right) \\ $$
Answered by $@ty@m last updated on 01/Jul/18
$${Eqn}.\:{of}\:{line}\:{passing}\:{through} \\ $$$${A}\:\&\:{B}\:{is}: \\ $$$${y}'−\mathrm{4}=\frac{\mathrm{1}}{{x}}\left({x}'−\mathrm{2}{x}\right) \\ $$$$\Rightarrow{x}\left({y}'−\mathrm{4}\right)=\left({x}'−\mathrm{2}{x}\right)\:...\left(\mathrm{1}\right) \\ $$$${It}\:{passes}\:{through}\:{C} \\ $$$$\therefore{x}\left(\mathrm{3}−\mathrm{4}\right)=\mathrm{4}−\mathrm{2}{x} \\ $$$$\Rightarrow−{x}=\mathrm{4}−\mathrm{2}{x} \\ $$$$\Rightarrow{x}=\mathrm{4} \\ $$$${Substituting}\:{in}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{4}\left({y}'−\mathrm{4}\right)={x}'−\mathrm{8} \\ $$$$\Rightarrow\mathrm{4}{y}^{'} −\mathrm{16}={x}'−\mathrm{8} \\ $$$$\Rightarrow{x}'−\mathrm{4}{y}^{'} +\mathrm{8}=\mathrm{0} \\ $$$$\:{or}\:{x}−\mathrm{4}{y}+\mathrm{8}=\mathrm{0} \\ $$