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  •  A moving line is <img alt="\mathrm{l \mathrm{x}+\mathrm{my}+\mathrm{n}=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bl%20%5Cmathrm%7Bx%7D&plus;%5Cmathrm%7Bmy%7

 A moving line is \mathrm{l \mathrm{x}+\mathrm{my}+\mathrm{n}=0}, where \mathrm{I,m,n}  are connected by the relations \mathrm{\mathrm{a} l+\mathrm{bm}+\mathrm{cn}=0,\ \&\ \mathrm{a}, \mathrm{b}, \mathrm{c}} are constants \mathrm{c\neq 0}. The line passes through a fixed point :

 

Option: 1

\mathrm{\left(\frac{c}{a}, \frac{c}{b}\right)}


Option: 2

\mathrm{\left(\frac{a}{c}, \frac{b}{c}\right)}


Option: 3

\mathrm{(a c, b c)}


Option: 4

\mathrm{(a b, c)}


Answers (1)

best_answer

The moving line is \mathrm{l \mathrm{x}+\mathrm{my}+\mathrm{n}=0 \ \ \ \ \ ........\left ( 1 \right )}
And \mathrm{\mathrm{a} l+\mathrm{bm}+\mathrm{cn}=0 \ \ \ \ \ ........\left ( 2 \right )}
Substituting the value of n from \mathrm{ ( 2 )} in \mathrm{ ( 1 )}, we have
\mathrm{ \begin{aligned} & \mathrm{c}(\mathrm{l} \mathrm{x}+\mathrm{my})-(\mathrm{a} l+\mathrm{bm})=0 \\ & \text { Or } \quad\left(\mathrm{x}-\frac{\mathrm{a}}{\mathrm{c}}\right)+\left(\frac{\mathrm{m}}{\mathrm{l}}\right)\left(\mathrm{y}-\frac{\mathrm{b}}{\mathrm{c}}\right)=0 \end{aligned}},
which is of the form \mathrm{P+\lambda Q=0 }    \mathrm{.........\left ( 3 \right ) }
\mathrm{\therefore } \mathrm{\left ( 3 \right ) }   is a line passing through the point of intersection of lines \mathrm{x-a / c=0 \text { and } y-b / c=0}
Hence the given line passes through the point \mathrm{(a/c, b/c)}, which is a fixed point since \mathrm{a, b, c} are constants.


 

Posted by

manish painkra

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