If three lines whose equations are $y = m_{1} x + c_{1}, y = m_{2} + c_{2} a n d y = m_{3} x + c_{3}$ 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

3 lines are concurrent if they pass through a common point i.e POI of any 2 lines on $3^{r d}$ line.

It is given that lines $y = m_{1} x + c_{1}$ __(1)

$y = m_{2} + c_{2}$ __(2)

$y = m_{3} x + c_{3}$ __(3) are concurrent

so, finding point of intersection of lines (1) & (3) subtracting (1) from (2)

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

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

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

Putting value of $x$ in equation (1)

$y = m_{1} x = 9$

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

So, POI of line (1) & (2) is $(\frac{c_{1} - c_{2}}{m_{2} - m_{1}}, \frac{m_{1} (c_{1} - c_{2})}{m_{2} m_{1}} + c_{1})$

Since 3 lines are concurrent point $(\frac{c_{1} - c_{2}}{m_{2} - m_{1}}, \frac{m_{1} (c_{1} - c_{2})}{m_{2} - m_{1}} + c_{1})$ will satisfy the equation of $3^{r d}$ line

Now putting $x = \frac{c_{1} - c_{2}}{m_{2} - m_{1}} & y = \frac{m_{1} (c_{1} - c_{2})}{m_{2} - m_{1}} + c_{1}$ in equation (3)

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

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

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

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

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

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

Hence proved

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Similar questions

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

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

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

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

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

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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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