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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
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).
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.
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