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If three lines whose equations are $y = m_1, x+ c_1, y = m_2, x+c_2$ and $y = m_3, x+c_3$ are concurrent, then show that $m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$

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  • If three lines may intersect at one point, then the point of intersection of lines (1) and (3) will also satisfy line (2).
The equation of the given lines are :
$ \qquad y = m_1x+c_1$------(1)
$ \qquad y = m_2x+c_2$--------(2)
$ \qquad y=m_3x+c_3$-------(3)
Subtract equation (1) and (2) we get,
$ 0= (m_2-m_1)x+(c_2-c_1)$
$ \Rightarrow x = \large\frac{c_2-c_1}{m_1-m_2}$
Substituting the value of $x$ in (1) we get,
$ y = m_1 \bigg( \large\frac{c_2-c_1}{m_1-m_2} \bigg) $$+c_1$
$ \Rightarrow y = \large\frac{m_1c_2-m_1c_1}{m_1-m_2}$$+c_1$
$ y =\large\frac{m_1c_2-m_1c_1+m_1c_1-m_2c_2}{m_1-m_2}$
$ \Rightarrow y = \large\frac{m_1c_2-m_2c_1}{m_1-m_2}$
Hence the point of intersection is $ \bigg( \large\frac{c_2-c_1}{m_1-m_2}$$, \large\frac{m_1c_2-m_2c_1}{m_1-m_2} \bigg)$
Since the line (1), (2) and (3) are concurrent, the above point of intersection should also satisfy the equation (3).
Hence substituting for $x$ and $y$ we get,
$ \large\frac{m_1c_2-m_2c_1}{m_1-m_2}$$ = m_3 \bigg( \large\frac{c_2-c_1}{m_1-m_2} \bigg)$$+c_3$
On simplifying we get,
$\large\frac{m_1c_2-m_2c_1}{m_1-m_2}$$ = \large\frac{m_3c_2-m_3c_1+c_3m_1-c_3m_2}{m_1-m_2}$
$ \Rightarrow m_1c_2-m_2c_1-m_3c_2-c_3m_1+m_2c_3=0$
$ \Rightarrow m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$
Hence proved.
answered May 20, 2014 by thanvigandhi_1

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