Question

If $a,b,c$ are in A.P., then $ax+by+c=0$ will always pass through a fixed point whose coordinates are

• A. $(1,-2)$
• B. $(-1,2)$
• C. $(1,2)$
• D. $(-1,-2)$

Answer 1

The correct option is A $(1,-2)$

Because $a,b,c$ are in AP, therefore $2b=a+c$.

Putting the value of $b$ in $ax+by+c=0$, we get
$ax+\frac{a+c}{2}y+c=0$
or, $a(2x+y)+c(y+2)=0$
or, $(2x+y)+\left[\frac{c}{a}\right](y+2)=0$
This equation represents a family of straight lines passing through the intersection of

$2x+y=0$ and $y+2=0$, i.e. $(1,-2)$.