Question

If $a,b,c$ are in A.P. and line $ax+by+c=0$ always passes through a fixed point $(h,k)$, then the equation of straight line perpendicular to the line $3x+y-2=0$ and passing through $(h,k)$is

• A. $x-3y+5=0$
• B. $x-3y-7=0$
• C. $x-3y=0$
• D. $x-3y-2=0$

Answer 1

The correct option is B $x-3y-7=0$

Given : $a,b,c$ are in A.P., so
$b=a+d,c=a+2d$
Now, putting this in the equation of straight line
$ax+by+c=0\Rightarrow ax+(a+d)y+(a+2d)=0\Rightarrow a(x+y+1)+d(y+2)=0\Rightarrow (y+2)+\frac{a}{d}(x+y+1)=0$
So it will always pass through
$y+2=0,x+y+1=0\Rightarrow y=-2,x=1$
Slope of the line $3x+y-2=0$
$m=-3$
Slope of line perpendicular is
$m^{\prime }=\frac{1}{3}$
Therefore, the required equation of line is
$y+2=\frac{1}{3}(x-1)$
$\therefore x-3y-7=0$