Question

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

• A. $y-x=\frac{7}{4}$
• B. $y-x=\frac{-7}{4}$
• C. $y-x=\frac{1}{4}$
• D. $y-x=\frac{-1}{4}$

Answer 1

The correct option is A $y-x=\frac{7}{4}$

$4acx+y(ab+bc+ca-abc)+abc=0$
Dividing by abc, we get $\frac{4}{b}x+\frac{3}{b}y+1-y=0\{\because \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\}$
$\Rightarrow \frac{1}{b}(4x+3y)+1-y=0$
Lines are concurrent at point of intersection of lines 4x+3y=0 and 1-y=0 or $(-\frac{3}{4},1)$
Hence required line is $y-x=\frac{7}{4}$