The equation of the straight line through the intersection of line $2 x + y = 1$ and $3 x + 2 y = 5$ and passes through the origin is

7 x + 3 y = 0

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7 x - y = 0

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3 x + 2 y = 0

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3 x + y = 0

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Solution

The correct option is A $7 x + 3 y = 0$
Given:
$L_{1} : 2 x + y - 1 = 0, L_{2} : 3 x + 2 y - 5 = 0$
The family of line through the intersection of $L_{1}$ and $L_{2}$ is
$L_{1} + λ L_{2} = 0$
So,
$(2 x + y - 1) + λ (3 x + 2 y - 5) = 0 \dots (1)$
Since, it passes through $(0, 0)$ , we get
$- 1 - 5 λ = 0 \Rightarrow λ = - \frac{1}{5}$

Putting the value of $λ$ in the equation $(1)$ , we get
$2 x + y - 1 - \frac{1}{5} (3 x + 2 y - 5) = 0$
$\Rightarrow \frac{7}{5} x + \frac{3}{5} y = 0$
$∴ 7 x + 3 y = 0$

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