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 $m_{1} (c_{2} - c_{1}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2})$ is equal to

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Solution

The correct option is B $0$
Given equation of lines are
$y = m_{1} x + c_{1}$ ...(i)
$y = m_{2} x + c_{2}$ ...(ii)
$y = m_{3} x + c_{3}$ ...(iii)

On solving Eqs. (i) and (ii), we get

$x = \frac{c_{1} - c_{2}}{m_{1} - m_{2}}$ and $y = \frac{m_{1} c_{2} - m_{2} c_{1}}{m_{1} - m_{2}}$

On puting the values of $x$ and $y$ in Eq. (iii), we get

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

$\Rightarrow \frac{m_{1} c_{2} - m_{2} c_{1}}{m_{1} - m_{2}} = \frac{m_{2} c_{2} + m_{3} c_{1} - m_{1} c_{3} - m_{2} c_{3}}{m_{1} - m_{2}}$

$\Rightarrow m_{1} c_{2} - m_{2} c_{1} = m_{3} c_{2} - m_{3} c_{1} + m_{1} c_{3} - m_{2} c_{3}$

$\Rightarrow m_{1} c_{2} - m_{2} c_{1} - m_{3} c_{2} + m_{3} c_{1} - m_{1} c_{3} + m_{2} c_{3} = 0$

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

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Similar questions

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

y = m_{3} + c_{3}

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

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.

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