statement 1:
alone is not sufficient, as nothing is known about the slopes. infinite lines with various slopes can pass through point (5,1)

statement 2:
alone is not sufficient. nothing can be done with intercepts alone.


taking together
plug (5,1) in the equation of both the lines

y = mx + c

5 = m1 + c1 .....equation for N
5 = m2 + c2 .....equation for P

m1 = 5 - c1
m2 = 5 - c2

if given c1 > c2 (intercept of N > intercept of P)

take all three conditions :
1) c1, c2 positive [let c1=2, c1=1]

m1 = 5 - c1 = 3
m2 = 5 - c2 = 4

m2 > m1



2) c1, c2 negative [let c1=-1, c1=-2]

m1 = 5 - c1 = 5 + 1 = 6
m2 = 5 - c2 = 5 + 2 = 7

m2 > m1

3) c1 positive, c2 negative [let c1=2, c1=-1]

m1 = 5 - c1 = 5 - 2 = 3
m2 = 5 - c2 = 5 + 1 = 6

m2 > m1

case 4 where c1 negative, c2 positive cannot be taken as that would make c1 < c2, hence not fulfilling condition in statement 2






so m2 > m1

hence both together sufficient

answer c

Hi all,
for sure i agree that the answer on this problem is C and i solved the same way as Mom4MBA, but with very little correction that in no way change the answer
plug (5,1) in the equation of both the lines

y = mx + c

5 = m1 + c1 .....equation for N
5 = m2 + c2 .....equation for P
here coordinates of the point where two lines intersect are x=5 and y=1 (5.1)
so the resulting equations must looks like
1=m1*5+c1
1=m2*5+c2
in this problem the reversed coordinates does not matter, but in others it is still useful

Another way of thinking of the problem

1) - gives no slope info so BCE
2) y intercept is not helpful for slope without another point.

think of drawing this - if both y intercepts are positive - then n the slope is negative and the slope of n is smaller (a larger negative). If both y intercepts are negative then both slopes are positive and n has the smaller slope. if one is negative and one is positive then the slope of n is negative and the slope of P is positive. So each time slope of p is bigger. Thus it will work.

OA is C