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

The equation of lines are

$y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and $y = m_{3} x + c_{3}$

i.e. $- m_{1} x + y - c_{1} = 0$

$- m_{2} x + y - c_{2} = 0$

and $- m_{3} x + y - c_{3} = 0$

We know that three lines are concurrent if

$∣ ∣ ∣ ∣ \begin{matrix} - m_{1} & 1 & - c_{1} - m_{2} & 1 & - c_{2} - m_{3} & 1 & - c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

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

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

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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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