For line l to intersect line m at point (2, 1/2), line m must have the point (2,1/2) on its graph. It is implied that line l already has the point (2, 1/2). If line m does not have it, then there will be no intersection at that specific point.
Try checking every choice to see if it has point (2,1/2) on the graph.
Note that (x,y) = (2,1/2); check by using x=2 and y=1/2.
For A,
2x = y/2
2(2) = (1/2) / 2
4 ≠ 1/4
Cannot be choice A as it results in a false equation; this choice will not go through (2,1/2)
For B:
2y = 3 - x
2(1/2) = 3 - 2
1 = 1
This is a true equation so the point (2,1/2) is on the graph of 2y=3-x. This means that if this is the equation for line m, then line m will have a point at (2,1/2) and therefore intersect with line l. Therefore, B is the answer.
The rest of the choices are false as shown:
For C:
2x + 4y = 8
2(2) + 4(1/2) = 8
4 + 2 = 8
6 ≠ 8
Cannot be choice C as it results in a false equation; this choice will not go through (2,1/2)
For D:
y = 4 - (5/4)x
1/2 = 4 - (5/4)(2)
1/2 = 4 - 5/2
1/2 = 8/2 - 5/2
1/2≠ 3/2
Cannot be choice D as it results in a false equation; this choice will not go through (2,1/2)
