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# If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1(c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0.

Q : 10       If three lines whose equations are  $y=m_1x+c_1,y=m_2x+c_2$ and   $y=m_3x+c_3$ 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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Concurrent lines means they all intersect at the same point
Now, given equation of lines are
$y=m_1x+c_1 \ \ \ \ \ \ \ \ \ \ \ -(i)$
$y=m_2x+c_2 \ \ \ \ \ \ \ \ \ \ \ -(ii)$
$y=m_3x+c_3 \ \ \ \ \ \ \ \ \ \ \ -(iii)$
Point of intersection  of equation (i) and (ii)  $\left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right )$

Now, lines are concurrent which means point $\left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right )$  also satisfy equation (iii)
Therefore,

$\frac{m_1c_2-m_2c_1}{m_1-m_2}=m_3.\left ( \frac{c_2-c_1}{m_1-m_2} \right )+c_3$

$m_1c_2-m_2c_1= m_3(c_2-c_1)+c_3(m_1-m_2)$

$m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$

Hence proved

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