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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.
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Solution
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
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