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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Solution

The three lines are

$y = m_{1} x + c_{1} . . . (1)$

$y = m_{2} x + c_{2} . . . (2)$

$y = m_{3} x + c_{3} . . . (3)$

Collinear or they meet at a point only when they have common point of intersection

Solving (1) and (2) for x and y

$m_{1} x + c_{1} = m_{2} x + c_{2}$

$x (m_{1} - m_{2}) = c_{2} - c_{1}$

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

$\Rightarrow y = m_{1} x + c_{1}$

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

$= m_{1} c_{2} - m_{1} c_{1} + m_{1} c_{1} - m_{2} c_{1}$

Putting x and y in (3)

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

$m_{1}^{2} c_{2} - m_{1} m_{2} c_{2} - m_{1} m_{2} c_{1} + m_{2}^{2} c_{1}$

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

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

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Similar questions

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

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

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_{3} x + c_{3}

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

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

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

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

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