Line n contains points (2, 4) and (-4, -2). Find the distance between line n and point B (3, 1).

Answer 1

first find line m in the form y = m x + b
m = (-2-4)/(-4 -2) = 6/6 = 1
so
y = 1 x + b
4 = 2 + b
b = 2
so
y = x + 2
now we want a line perpendicular to that one through (3,1)
slope = -1/1 = -1
y = -x + b
1 = -3 + b
b = 4
y = -x + 4
now were does that perpendicular hit the first line?
y = -x + 4
y = x + 2
2 y = 6
y = 3
x = 1
so
we want distance from (1,3) to (3,1)
d = sqrt (-2^2 + 2^2) = sqt(4+4) = 2 sqrt 2

Answer 2

As shown above, the line is
y = x + 2
x-y+2 = 0
So, the distance from (3,1) to the line is
|3-1+2|/√(1+1) = 4/√2 = 2√2