Question

If $a,b,c$ are in GP, then find he area of the triangle formed by the lines ax+by+c=$0$ with the coordinates axes.

Answer 1

The point of intersection of the line $ax+by+c=0$ with the coordinate axis are $(\frac{-c}{a},0)$ and $(0,\frac{-c}{b})$

So, the vertices of the triangle are $(0,0)$,$(\frac{-c}{a},0)$ and $(0,\frac{-c}{b})$

Let $A$ be the area of the required triangle

$=\frac{1}{2}|\frac{-c}{a}\times \frac{-c}{b}|=\frac{1}{2}\left|\frac{c^2}{ab}\right|$

It is given that $a,b$ and $c$ are in G.P

$\therefore b^2=ac$

$\Rightarrow A=\frac{1}{2}\left|\frac{b^4}{a^2\times ab}\right|=\frac{1}{2}{\left|\frac{b}{a}\right|}^3$sq.units.