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

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Solution

$m_{1} x - y + a_{1} = 0 m_{2} x - y + a_{2} = 0 m_{3} x - y + a_{3} = 0$

Lines meet in a point if $∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

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

Simplifying using properties of determinant

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

Expanding along $C_{2}$

$0 + 0 - 1 {(m_{1} - m_{2}) {(a}_{2} - a_{3}) - (m_{2} - m_{3}) (a_{1} - a_{2})} = 0 m_{1} a_{2} - m_{1} a_{3} - m_{2} a_{2} + m_{2} a_{3} - m_{2} a_{1} + m_{2} a_{2} + m_{3} a_{1} - m_{3} a_{2} = 0 m_{1} a_{2} - m_{1} a_{3} + m_{2} a_{3} - m_{2} a_{1} + m_{3} a_{1} - m_{3} a_{2} = 0 m_{1} (a_{2} - a_{3}) + m_{2} (a_{3} - a_{1}) + m_{3} (a_{1} - a_{2}) = 0$

is the required condition.

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Similar questions

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.

x_{1}, x_{2}, x_{3}

A_{1}, A_{2}, A_{3}

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

y = m_{3} x

2 x - y + 3 = 0

m_{1}, m_{2}, m_{3}

A . P .

x_{1}, x_{2}, x_{3}

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

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

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

y = m_{1} x + \frac{a}{m_{1}}

y = m_{2} x + \frac{a}{m_{2}}

y = m_{3} x + \frac{a}{m_{3}}

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

y = m_{3} x

x + y = 1,

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