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
Co-ordinate Geomatry
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Source: Beat The GMAT — Data Sufficiency |
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clock60
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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
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
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tomada
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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
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