The slope of EF is −5/2 .

1 answer:

4 0

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

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