Question

If $a,c,b$ are three terms of a geometric progression, then the line $ax+by+c=0$

• A. has a fixed direction
• B. always passes through a fixed point
• C. forms a triangle with the coordinate axes whose area is constant
• D. always cuts intercepts on the coordinate axes such that their sum is zero

Answer 1

The correct option is C

forms a triangle with the coordinate axes whose area is constant

$ax+by+c=0$
$\Rightarrow ax+by=-c$
$\frac{x}{\left(\frac{-c}{a}\right)}+\frac{y}{\left(\frac{-c}{b}\right)}=1$
Here, $x-\text{intercept}=\frac{-c}{a},y-\text{intercept}=\frac{-c}{b}$


As $a,c,b$ are in G.P., $c^2=ab$
Area of $△AOB$ is $\frac{1}{2}\times OA\times OB$
$=\frac{1}{2}\times \left(\frac{-c}{a}\right)\times \left(\frac{-c}{b}\right)$
$=\frac{c^2}{2ab}=\frac{ab}{2ab}$
$=\frac{1}{2}$ sq. units which is a constant.