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If $a, c, b$ are in G.P., then the area of the triangle formed by the lines $a x+b y+c=0$ with the coordinates axes is equal to
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$\frac{1}{2}$
Area of the triangle $=\frac{1}{2}(\mathrm{x}$ intercept $) \times(\mathrm{y}$ intercept $)$
$=\frac{1}{2}\left(-\frac{\mathrm{c}}{\mathrm{a}}\right)\left(-\frac{\mathrm{c}}{\mathrm{b}}\right)=\frac{1}{2} \frac{\mathrm{c}^{2}}{\mathrm{ab}}=\frac{1}{2} \mathrm{unit}$
$\left[\because \mathrm{a}, \mathrm{c}, \mathrm{b}\right.$ are in $\left.\mathrm{G} . \mathrm{P} . \Rightarrow \mathrm{c}^{2}=\mathrm{ab}\right]$
$=\frac{1}{2}\left(-\frac{\mathrm{c}}{\mathrm{a}}\right)\left(-\frac{\mathrm{c}}{\mathrm{b}}\right)=\frac{1}{2} \frac{\mathrm{c}^{2}}{\mathrm{ab}}=\frac{1}{2} \mathrm{unit}$
$\left[\because \mathrm{a}, \mathrm{c}, \mathrm{b}\right.$ are in $\left.\mathrm{G} . \mathrm{P} . \Rightarrow \mathrm{c}^{2}=\mathrm{ab}\right]$
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