The slope of ef is −5/2 . which segments are perpendicular to ef? select all the right answers 1. jk , where j is at (3, −2) and k is at (5, −7) 2. gh , where g is at (6, 7) and h is at (4, 12) 3. lm , where l is at (1, 9) and m is at (6, 11) 4. np , where n is at (−3, 4) and p is at (−8, 2)

We know that
if two lines are perpendicular
then 
m1*m2=-1
where 
m1 and m2 are the slopes of the lines

so
if the slope of EF is −5/2
then 
the slope of  the segment  perpendicular to EF is equal to
m1*m2=-1> m2=-1/m1> m2=-1/(-5/2)> m2=2/5

case N 1) JK , where J is at (3, −2) and K is at (5, −7)

find the slope JK
m=(y2-y1)/x2-x1)> m=(-7+2)/(5-3)> m=-5/2
-5/2 is not equal to 2/5> JK is not perpendicular to EF

case N 2) GH , where G is at (6, 7) and H is at (4, 12)
find the slope GH
m=(y2-y1)/x2-x1)> m=(12-7)/(4-6)> m=5/-2> m=-5/2
-5/2 is not equal to 2/5> GH is not perpendicular to EF

case N 3) LM , where L is at (1, 9) and M is at (6, 11)
find the slope LM
m=(y2-y1)/x2-x1)> m=(11-9)/(6-1)> m=2/5
2/5 is equal to 2/5> LM is perpendicular to EF

case N 4) NP , where N is at (−3, 4) and P is at (−8, 2)
find the slope NP
m=(y2-y1)/x2-x1)> m=(2-4)/(-8+3)> m=-2/-5> m=2/5
2/5 is equal to 2/5> NP is perpendicular to EF

the answer is
The segments perpendicular to EF are LM and NP

The third box is a function