The equation of the line passing through the origin and having slopes $3$ and $- \frac{1}{3}$ is

3 y^{2} + 8 x y - 3 x^{2} = 0

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3 x^{2} + 8 x y - 3 y^{2} = 0

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3 y^{2} - 8 x y + 3 x^{2} = 0

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3 x^{2} + 8 x y - 3 y^{2} = 0

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Solution

The correct option is A $3 y^{2} + 8 x y - 3 x^{2} = 0$
General eqn. of line passing through origin having slope $m$ ,
$\Rightarrow y = m x$
$\Rightarrow$ Eqn. of 1 st line having slope 3 is $y = 3 x$
$\Rightarrow y - 3 x = 0 \to (1)$

Eqn of line having slope $- \frac{1}{3}$ is,

$= 7 - \frac{1}{3} = \frac{y}{x}$
$\Rightarrow 3 y = - x \to (1)$

$\Rightarrow 3 y + x = 0 \to (2)$
On multiplying eqn. $(1) & (2)$ we get,

$(y - 3 x) (3 y + x) = 0$

$\Rightarrow 3 y^{2} - x y + 9 x y - 3 x^{2} = 0$
$= 3 y^{2} + 8 x y - 3 x^{2} = 0$
If is the required equation of lines.

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3 x^{2} - 8 x y - 3 y^{2} = 0

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and

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3 x^{2} - 8 x y - 3 y^{2} = 0

and

x - 2 y = 3

represents the sides of a triangle which is

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The equation of the line passing through the origin and having slopes 3 and −13 is

The equation of the line passing through the origin and having slopes $3$ and $- \frac{1}{3}$ is