Given: lines *y* = 3*x* + 1, *y* = -2*x* + 5, and a third line L_{1}: *y* = *mx* + *c* with *m* > 0 and *c* a constant. L_{1} moves parallel to itself intersecting the two given lines in points P_{1} and P_{2} respectively. The locus of mid-points of segment P_{1}P_{2} is line L_{2}. Find the slope of L_{1} so that L_{2} has a slope that is undefined.

- 1/3
- 1/2
- 2/3
- 1
- 3/2

Here’s the scene. If we can find two points on L3 → 5 → P

_{1}, we can get its slope*m*. One point, clearly, is where the two given lines intersect. Call it P_{3}. There, P_{1}= P_{2}(= P_{3}) and the midpoint is P_{3}also. So:*x*+ 1 = -2

*x*+ 5

*x*= 4

*x*= 4/5 and

*y*= 17/5.

_{3}is (4/5, 17/5).

Since we want LP and P . If we can find the coordinates of P

_{2}to be vertical, another easy point to look at on L_{2}is the intersection of L_{2}with the*x*-axis, the point (4/5, 0), which is the midpoint between two as-yet-unknown points_{1}(

*x*

_{1}, 3

*x*

_{1}+ 1)

_{2}(

*x*

_{2}, -2

*x*

_{2}+ 5)

_{2}, then we’ll have the coordinates of two points on L_{1}and can find the slope of the line.
Using the midpoint formula, we get two equations in

*x*_{1}and*x*_{2}:(

*x*_{1}+*x*_{2})/2 = 4/5(3

*x*_{1}+ 1 -2*x*_{2}+ 5)/2 = 0Solving these two simultaneous equations (some algebra here), we get

*x*_{1}= -14/25 and*x*_{2}= 54/25. Now we can find our points P_{1}on*y*= 3*x*+ 1 and P_{2}on*y*= -2*x*+ 5:P

_{1}= (-14/25, -17/25)P

_{2}= (54/25, 17/25)Using the slope formula, we find the slope of L

_{1}to be:*m*= [17/25 – (-17/25)] / [54/25 – (-14/25)] = 34/68 = 1/2.

The answer is (b). If the slope of L

_{1}is 1/2, then the slope of L_{2}is undefined.