Let `l` be the line belonging to the family of straight lines `(a + 2b)x+ (a - 3b)y +a-8b = 0, a, b in R`, which is farthest from the point `(2, 2),` then area enclosed by the line `L` and the coordinate axes is

Options

A. x+4y+7=0
B. 2x+3y+4=0
C. 4x-y-6=0
D. none of these

Answer

A

Explanation

Given lines (x+y+1) +b(2x-3y-8) = 0 are concurrent at the point of intersection of the line x+y+1=0 and 2x-3y-8=0, which is (1, -2). Now, the line through A(1, -2), which is farthest from the point B(2, 2), is perpendicular to AB. Now, the slope of AB is 4. Then the required line is y+2 =-(1/4)(x-1) or x+4y=0.