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 \[\left( - \frac{c}{a}, 0 \right) \text { and } \left( 0, - \frac{c}{b} \right)\].

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

Let A be the area of the required triangle.

\[A = \frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ \frac{- c}{a} & 0 & 1 \\ 0 & \frac{- c}{b} & 1\end{vmatrix}\]

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

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