Consider a family of straight lines (x + y) + lambda(2x - y + 1) = 0. Find the equation of the straight line belonging to this family that is farthest from (1, -3).

Kindly please don't refer me the link to a similar question that ha s already been answered as I tried that method but I got a wrong answer.

Thank You.

Dear student,

Please find below the solution to the asked query:

Consider a family of staright lines.    x+y+λ2x-y+1=0This family consists of two lines:        x+y=0   and  2x-y+1=0Add the two equations to get,    3x+1=0        3x=-1         x=-13Put this value in x+y=0    -13+y=0         y=13So we have x,y=A-13,13Let the given point be B1,-3Note that the required line is perpendicular to line AB.The slope of the line AB is,    m1=-3-131+13=-9-13+1=-104=-52So the slope of the required line is,   m2=-1m1 =-1-52=25Use point slope form to find the required equation of line.    y-y1=m2x-x1    y-13=25x+13    3y-1=253x+1   5 3y-1=23x+1   15y-5=6x+2  15y-6x-7=0So the required equation of line is,  15y-6x-7=0

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