If $a, c, b$ are in G.P. and the area formed by the coordinate axes and the line $a x + b y + c = 0$ is $A sq. units,$ then the value of $4 A$ is

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Solution

$a x + b y + c = 0 \Rightarrow \frac{x}{- c / a} + \frac{y}{- c / b} = 1$
Therefore, $x -$ intercept $= - \frac{c}{a}$ and $y -$ intercept $= - \frac{c}{b}$
So, area of the triangle,
$A = \frac{1}{2} \times ∣ ∣ ∣ (- \frac{c}{a}) \times (- \frac{c}{b}) ∣ ∣ ∣ \Rightarrow A = \frac{c^{2}}{2 a b}$
As $a, c, b$ are in G.P., so $c^{2} = a b$
$∴ A = \frac{a b}{2 a b} = \frac{1}{2} \Rightarrow 4 A = 2$

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If a,c,b are in G.P. and the area formed by the coordinate axes and the line ax+by+c=0 is A sq. units, then the value of 4A is

ax+by+c=0⇒x−c/a+y−c/b=1 Therefore, x−intercept =−ca and y−intercept =−cb So, area of the triangle, A=12×∣∣∣(−ca)×(−cb)∣∣∣⇒A=c22ab As a,c,b are in G.P., so c2=ab ∴A=ab2ab=12⇒4A=2

If $a, c, b$ are in G.P. and the area formed by the coordinate axes and the line $a x + b y + c = 0$ is $A sq. units,$ then the value of $4 A$ is