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Question Number 113644 by AbhishekBasnet last updated on 14/Sep/20

Answered by 1549442205PVT last updated on 14/Sep/20

The intersection point of two lines  x−y+1=0 and 2x−y−1=0 is roots  of the system: { ((x−y+1=0)),((2x−y−1=0)) :}  Substracting two  above equations  we get x−2=0⇒x=2,y=3.Replace   this value into the equation   x+3y−11=0 we see they satisfy.That  shows that three above lines are  concurren at M(2,3)(Q.E.D)

$$\mathrm{The}\:\mathrm{intersection}\:\mathrm{point}\:\mathrm{of}\:\mathrm{two}\:\mathrm{lines} \\ $$$$\mathrm{x}−\mathrm{y}+\mathrm{1}=\mathrm{0}\:\mathrm{and}\:\mathrm{2x}−\mathrm{y}−\mathrm{1}=\mathrm{0}\:\mathrm{is}\:\mathrm{roots} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{system}:\begin{cases}{\mathrm{x}−\mathrm{y}+\mathrm{1}=\mathrm{0}}\\{\mathrm{2x}−\mathrm{y}−\mathrm{1}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{Substracting}\:\mathrm{two}\:\:\mathrm{above}\:\mathrm{equations} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{x}−\mathrm{2}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{2},\mathrm{y}=\mathrm{3}.\mathrm{Replace}\: \\ $$$$\mathrm{this}\:\mathrm{value}\:\mathrm{into}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{x}+\mathrm{3y}−\mathrm{11}=\mathrm{0}\:\mathrm{we}\:\mathrm{see}\:\mathrm{they}\:\mathrm{satisfy}.\mathrm{That} \\ $$$$\mathrm{shows}\:\mathrm{that}\:\mathrm{three}\:\mathrm{above}\:\mathrm{lines}\:\mathrm{are} \\ $$$$\mathrm{concurren}\:\mathrm{at}\:\mathrm{M}\left(\mathrm{2},\mathrm{3}\right)\left(\boldsymbol{\mathrm{Q}}.\boldsymbol{\mathrm{E}}.\boldsymbol{\mathrm{D}}\right) \\ $$

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