Question

If a, b, c are in G.P. write the area of the triangle formed by the line 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$\left(-\frac{c}{a},0\right)\mathrm{and}\left(0,-\frac{c}{b}\right)$.

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

Let A be the area of the required triangle.

$$\begin{gathered} A=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ \frac{-c}{a}& 0& 1\\ 0& \frac{-c}{b}& 1\end{array}\right| \\ A=\frac{1}{2}\left|-\frac{c}{a}\times \frac{-c}{b}\right|=\frac{1}{2}\left|\frac{c^2}{ab}\right| \end{gathered}$$

It is given that a, b and c are in GP.

$\therefore b^2=ac$

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