$\begin{matrix} A & B A_{1} & B_{1} A_{2} & B_{2} \end{matrix}$

If A and B are in inverse proportion, then which of the following is true?

\frac{A_{1}}{B_{1}} = \frac{B_{2}}{A_{2}}

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\frac{A_{1}}{B_{2}} = \frac{B_{2}}{A_{1}}

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A_{1} \times B_{1} = A_{2} \times B_{2}

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A_{1} \times A_{2} = B_{1} \times B_{2}

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Solution

The correct option is C $A_{1} \times B_{1} = A_{2} \times B_{2}$
If A and B are in inverse proportion,

$\frac{A_{1}}{A_{2}} = \frac{B_{2}}{B_{1}}$

$\Rightarrow A_{1} \times B_{1} = A_{2} \times B_{2}$

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Fill in the blanks:

a : b and c : d are in inverse proportion, then a : b = ...........

\frac{a^{2}}{b^{2}} = \frac{a c}{b d}

Which of the following is true for inverse proportion?

\begin{matrix} A & B A_{1} & B_{1} A_{2} & B_{2} \end{matrix}

A_{1} B_{1}

A_{2} B_{2}

y^{2} = 4 a x

A_{1} A_{2}

B_{1} B_{2}

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