Question

If $a,b,c$ are in A.P., then the line $ax+by+c=0$ will always pass through a fixed point $(h,k)$, then the value of $h-k$ is

Answer 1

We know that
$2b=a+c$
Putting the value of $b$ in $ax+by+c=0$, we get
$ax+\left(\frac{a+c}{2}\right)y+c=0$
$\Rightarrow a(2x+y)+c(y+2)=0$
$\Rightarrow (2x+y)+\left(\frac{c}{a}\right)(y+2)=0$
This equation represents a family of straight lines passing through the intersection of $2x+y=0$ and $y+2=0$ which is $(1,-2)$
Hence, the value of $h-k=1+2=3$


Alternate solution:
We know that
$2b=a+c\Rightarrow a-2b+c=0\cdots \left(1\right)$
comparing above equation with $ax+by+c=0\cdots \left(2\right)$
Line always through point
$\Rightarrow x=1,y=-2$
$\therefore h-k=1+2=3$