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Given the system of straight lines a(2x+y-3) + b(3x+2y-5) = 0, the line of the system farthest from the point (4,-3) has the equation?


Options: a) 4x+11y-15=0


             b) 7x+y-8=0


             c) 4x+3y-7=0


             d) 3x-4y+1=0

3 years ago

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Answers : (2)

                                        

 Hi Suhas,


 


The given system represents a family of lines passing through the intersection of 2x+y-3=0 and 3x+2y-5=0


These two lines intersect at (1,1).


 


Now a line throug this system will be the farthest from the point (4,-3) if the line is perpendicular to the line joining (1,1) and (4,-3).


Slope of the line joining (4,-3) and (1,1) is -4/3.


So slope of the required line passing through (1,1) is 3/4.


 


So the eqn of the line will be 3x-4y+7=0 (there is no option in the question, please check your choice D).


 


Regards,


Ashwin (IIT Madras).

3 years ago
                                        

Hi Suhas,


 


Sorry. A silly mistake in the last step.


Eqn of line through (1,1) and having slope as 3/4 will be 3x-4y+1=0


So option (D) is correct.


 


Regards,


Ashwin (IIT Madras).

3 years ago

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