A line segment has endpoints j ( 2, 4 ) and l ( 6, 8 ). The point k is the midpoint of jl. What is an equation of a line perpendicular to jl and passing through k?

Answer 1

Slope of segment, a
= (y2-y1)/(x2-x1)
= (8-4)/(6-2)=1
Mid-point
=( (x1+x2)/2, (y1+y2)/2 )
=( (2+6)/2, (4+8)/2 )
= M(4,6)

Slope of line perpendicular to segment,a1
= -1/a
= -1/(1)
=-1

Equation of perpendicular line:
(y-y1)=a1(x-x1)
y-6 = -1(x-4)
or
y=-x + 10

Check that it goes through M
-4+10=6 ok.

Answer 2

y=-x+10

Answer 3

gdxgdx

Answer 4

y=x+2

Answer 5

Slope of segment, a
= (y2-y1)/(x2-x1)
= (8-4)/(6-2)=1
Mid-point
=( (x1+x2)/2, (y1+y2)/2 )
=( (2+6)/2, (4+8)/2 )
= M(4,6)

Slope of line perpendicular to segment,a1
= -1/a
= -1/(1)
=-1

Equation of perpendicular line:
(y-y1)=a1(x-x1)
y-6 = -1(x-4)
or
y=-x + 10

Check that it goes through M
-4+10=6 ok.