Based on the above, the segments that are perpendicular to EF are LM and NP.
Note that when two lines are perpendicular, we can say that;
M1 * M2 = -1 As M1 and M2 are known to be the slopes of the lines.
Therefore, when the the slope of EF is said to be −5/2, then one can say that the slope of the segment that is said to be perpendicular to EF will have to be equal to m1*m2=-1, m2=-1/m1, m2=-1/(-5/2) or m2=2/5.
Scenario one:
JK , if J is at (3, −2) and K is at (5, −7)
To find the slope JK, then
m=(y2-y1)/x2-x1)
m=(-7+2)/(5-3)
m=-5/2
-5/2 is not equal to 2/5
Therefore, JK is not perpendicular to EF
Scenario 2
Find GH , when G is at (6, 7) and H is at (4, 12)
To find the slope GH
m=(y2-y1)/x2-x1)
m=(12-7)/(4-6)
m=5/-2
m=-5/2
Since -5/2 is not equal to 2/5 then GH is not perpendicular to EF
Scenario 3:
Find LM , If L is at (1, 9) and M is at (6, 11)
To find the slope LM, then
m=(y2-y1)/x2-x1)
m=(11-9)/(6-1)
m=2/5
Since 2/5 is equal to 2/5
Then LM is perpendicular to EF
Scenario 4:
Find NP , if N is at (−3, 4) and P is at (−8, 2)
To find the slope NP, then
m=(y2-y1)/x2-x1)
m=(2-4)/(-8+3)
m=-2/-5
m=2/5
Since 2/5 is equal to 2/5.
Therefore, NP is perpendicular to EF
Based on the above calculations, the segments that are perpendicular to EF are LM and NP.
See correct format of question written below
The slope of EF is −5/2 .
Which segments are perpendicular to EF?
Select all the right answers please
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)
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