If the three lines $3y-2x=1$, $x+2y=2$ and $4x-6y=5$ are drawn in the plane, how many points will lie at the i

Question

If the three lines $3y-2x=1$, $x+2y=2$ and $4x-6y=5$ are drawn in the plane, how many points will lie at the intersection of at least two of the three lines?

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Reason: lines 3y-2x=1 and 4x-6y=5 have the same slope (2/3) but different intercepts, so they are parallel and do not meet. The other line x+2y=2 meets each of them in one distinct point, giving 2 intersection points.

If the three lines $3y-2x=1$, $x+2y=2$ and $4x-6y=5$ are drawn in the plane, how many points will lie at the intersection of at least two of the three lines?

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