The intercept form of the line $3 x - 4 y + 12 = 0$ is

\frac{x}{4} - \frac{y}{3} = 1

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\frac{x}{- 4} + \frac{y}{3} = 1

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\frac{x}{- 4} + \frac{y}{3} = 2

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\frac{x}{4} - \frac{y}{3} = 2

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Solution

The correct option is B $\frac{x}{- 4} + \frac{y}{3} = 1$
$3 x - 4 y + 12 = 0$
$\Rightarrow 3 x - 4 y = - 12$
$\Rightarrow \frac{x}{- 4} + \frac{y}{3} = 1$

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