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

Hope this information will clear your doubts about the topic.

If you have any more doubts, just ask here on the forum and our experts will try to help you out as soon as


  • 107
What are you looking for?