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  • Consider the family of lines <img alt="\mathrm{5 x+3 y-2+11(3 x-y-4)=0 \: and\: x-y+1+12(2 x-y-2)=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B5%20x&plus;3%20y-2&plu

Consider the family of lines \mathrm{5 x+3 y-2+11(3 x-y-4)=0 \: and\: x-y+1+12(2 x-y-2)=0}. Equation of a straight line that belong to both families is -
 

Option: 1

25 \mathrm{x}-62 \mathrm{y}+86=0

 


Option: 2

\mathrm{62 x-25 y+86=0}
 


Option: 3

\mathrm{25 x-62 y=86}
 


Option: 4

\mathrm{5 \mathrm{x}-2 \mathrm{y}-7=0}


Answers (1)

best_answer

Lines \mathrm{5 \mathrm{x}+3 \mathrm{y}-2+11(3 \mathrm{x}-\mathrm{y}-4)=0} are concurrent at \mathrm{(1,-1)} and lines \mathrm{\mathrm{x}-\mathrm{y}+1+12(2 x-y-2)=0} are concurrent at (3,4). Thus equation of line common to both family is,

\mathrm{ y-4=\frac{-1-4}{1-3}(x-3) }

\mathrm{ \text { i.e. } 5 x-2 y-7=0 }

Hence option 4 is correct.



 

Posted by

seema garhwal

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