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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Solution

$y = m_{1} x + c_{1} . . . . . . . . (i)$
$y = m_{2} x + c_{2} . . . . . . . . (i i)$
$y = m_{3} x + c_{3} . . . . . . . . . . (i i i)$
Subtracting $(i)$ and $(i i)$ ,we get
$x = \frac{c_{2} - c_{1}}{m_{1} - m_{2}}$
Putting in eqn(i), we get,
$y = \frac{m_{1} (c_{2} - c_{1})}{m_{1} - m_{2}} + c_{1}$
$y = \frac{m_{1} c_{2} - m_{2} c_{1}}{m_{1} - m_{2}}$
Putting it the eqn(iii), we get
$\frac{m_{1} c_{2} - m_{2} c_{1}}{m_{1} - m_{2}} = \frac{m_{3} (c_{2} - c_{1}) + m_{1} c_{3} - m_{2} c_{3}}{m_{1} - m_{2}}$
Rearranging, we get
$m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$

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Similar questions

If the three lines $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and $y = m_{3} x + c_{3}$ are concurrent then show that,

$m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$

y = m_{1} x + c_{1}, y = m_{2} x + c_{2}

y = m_{2} x + c_{2}

m_{1} (c_{2} - c_{1}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0

y = m_{1} x + c_{1}

y = m_{2} x + c_{2}

y = m_{3} x + c_{3}

Find the conditions that the straight lines $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and $y = m_{3} x + c_{3}$ may meet in a point.

y = m_{1} x + c_{1}

y = m_{2} x + c_{2}

y

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