Equation of line which is equally inclined to the axis and passes through a common points of family of lines `4acx + y(ab + bc + ca- abc) +abc = 0` (where `a, b, c> 0` are in `H.P.)` is

Options

A. `y - x = (7)/(4)`
B. `y - x =- (7)/(4)`
C. `y - x =(1)/(4)`
D. `y -x =- (1)/(4)`

Answer

A

Explanation

`4acx +y (ab +bc +ca- abc) +abc = 0`
Dividing by abc, we get` (4)/(b) x +y(3)/(b) +1 -y = 0 ( :' (2)/(b) =(1)/(a)+(1)/(c))`
`:. (1)/(b) (4x +3y) +1 -y = 0`
Lines are concurrent at point of intersesction of lines `4x +3y =0` and `1 - y = 0` or `(-3//4,1)`
Hence the required line in `y -x =(7)/(4)`