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(A) First,find the point of intersection of the lines $x - 3y + 1 = 0$ and $2x + 5y - 9 = 0$. Multiplying the first equation by $2$,we get $2x - 6y +

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$x = 2$

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(A) First,find the point of intersection of the lines $x - 3y + 1 = 0$ and $2x + 5y - 9 = 0$. Multiplying the first equation by $2$,we get $2x - 6y + 2 = 0$. Subtracting this from $2x + 5y - 9 = 0$,we get $11y - 11 = 0$,which implies $y = 1$. Substituting $y = 1$ into $x - 3y + 1 = 0$,we get $x - 3(1) + 1 = 0$,so $x = 2$. The point of intersection is $(2, 1)$. $A$ line with an infinite slope is a vertical line of the form $x = k$. Since the line passes through $(2, 1)$,the equation must be $x = 2$. The distance of the line $x = 2$ from the origin $(0, 0)$ is $|2| = 2 	ext{ units}$,which satisfies the given condition. Thus,the equation of the line is $x = 2$.

The straight line passing through the point of intersection of the straight lines $x - 3y + 1 = 0$ and $2x + 5y - 9 = 0$ and having infinite slope and at a distance of $2 \text{ units}$ from the origin,has the equation

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