stmt1,

L1:

2=m+C

1=2m+C

m=-1,C=3

Slope=1[not llr]

Suff

stmt2,

L1 slope=slope of y=x

L1 slope=1

L2 slope=slope of y=x+4

L2 slope=1

Slope=1[not llr]

Suff

Pick D

i am not dead sure in selango`s answer
to me the answer is B
my way
for two lines be perpendicular the product of their slopes must look like -k*(1/k)=-1 ( as sin90=1, and one slope must be -ve)
(1) i don`t see how to calculate the slope from 1 st, lines can have slope they want to.
(2) line 1 || y=x so it means that slope l1=1 (y=1*x)
line2 || y=x+4, y=1*x+4
so slope of line 2=1 ,from this we can conclude that slope of l1=slope of l2 and they can`t be perpendicular
B is my pick

Clock, you are correct. The answer is B. Only Statement 2 is sufficient.

clock60 wrote:i am not dead sure in selango`s answer
to me the answer is B
my way
for two lines be perpendicular the product of their slopes must look like -k*(1/k)=-1 ( as sin90=1, and one slope must be -ve)
(1) i don`t see how to calculate the slope from 1 st, lines can have slope they want to.
(2) line 1 || y=x so it means that slope l1=1 (y=1*x)
line2 || y=x+4, y=1*x+4
so slope of line 2=1 ,from this we can conclude that slope of l1=slope of l2 and they can`t be perpendicular
B is my pick