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oh it's an intercept given in st(2)

e
two parallel lines may have horizontal distance of 5 units and different slopes with one line crossing y-abscess on 2.5 and still these lines could be located anywhere in the X-Y space

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Per my understanding:

since the 2 lines are paralllel, the gradient/ slope of the 2 lines need to be same.

Linear line equation: y=Mx+b

From statement 1: line M: y=mx (since the line crosses the origin), but it gives no idea about the 2nd line.and horizontal distance between the 2 lines is 5

From statement 2: Line L: y=Mx+2.5.

In order to figure out the equation for the line, we need the value of M. M can be calculated as from the right angles triangle (base= 5.0, height = 2.5)

Hence both the statements are needed for getting the equation for both the lines.[/img]

according to you both statements taken together are Sufficient? Is this what you mean?
answer C is the wrong answer and you miss different positions of the triangle shaped figures in X-Y space; the triangles apexed on (0,0) could be positioned in infinite number of ways

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That is my understanding of it. But i might be wrong.