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If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0
If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0
Class 11 | Maths | Miscellaneous Exercise | NCERT | Straight Lines
If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0

It is provided that

\[y\text{ }=\text{ }{{m}_{1}}x\text{ }+\text{ }{{c}_{1}}~\ldots ..\text{ }\left( 1 \right)\]

\[y\text{ }=\text{ }m2x\text{ }+\text{ }c2\text{ }\ldots \text{ }..\text{ }\left( 2 \right)\]

\[y\text{ }=\text{ }m3x\text{ }+\text{ }c3\text{ }\ldots \text{ }..\text{ }\left( 3 \right)\]

 

By deducting condition (1) from (2) we get

 

\[0\text{ }=\text{ }\left( m2\text{ }\text{ }m1 \right)\text{ }x\text{ }+\text{ }\left( c2\text{ }\text{ }c1 \right)\]

\[\left( m1\text{ }\text{ }m2 \right)\text{ }x\text{ }=\text{ }c2\text{ }\text{ }c1\]

So we get

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 56

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 57

Taking out the normal terms

 

\[m1\text{ }\left( c2\text{ }\text{ }c3 \right)\text{ }+\text{ }m2\text{ }\left( c3\text{ }\text{ }c1 \right)\text{ }+\text{ }m3\text{ }\left( c1\text{ }\text{ }c2 \right)\text{ }=\text{ }0\]

Thus, \[m1\text{ }\left( c2\text{ }\text{ }c3 \right)\text{ }+\text{ }m2\text{ }\left( c3\text{ }\text{ }c1 \right)\text{ }+\text{ }m3\text{ }\left( c1\text{ }\text{ }c2 \right)\text{ }=\text{ }0.\]

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