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).

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).