If $a, b, c$ are in $G . P .$ then the line $a^{2} x + b^{2} y + a c = 0$ will always pass through the fixed point

(0, 1)

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(1, 0)

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(0, - 1)

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(1, - 1)

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Solution

The correct option is C $(0, - 1)$
$\begin{matrix} I f a, b, (a r e i n G P t h e n) b = a r c = a r^{2} o n p u t t i n g i n e q u a t i o n . a^{2} x + a^{2} r^{2} y + a^{2} r^{2} = 0 [x + r^{2} (y + 1) = 0] r^{2} = - \frac{x}{1 + y} F r o m t h i s x = 0 a n d y = - 1 w i l l b e t h e f i x e d p o i n t f r o m w h i c h e q u a t i o n w i l l p a s s \end{matrix}$

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