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

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