find the distance from the point (1,1) to the line y=3x+2 .
i did
y = 3x + 2
slope > 3
distance > -1/3
distance has slope -1/3 passes through the point (1,1)
y - y1 = m(x - x1) < point slope formula
y - 1 = (-1/3)(x-1)
y - 1 = -1/3x + 1/3
y = -1/3x + 1 1/3
y = -1/3x + 4/3
y=-1/3x + 4/3
y = 3x + 2
find the intersection point
-1/3x + 4/3 = 3x + 2
-x + 4 = 9x + 6
-10x = 2
x = -5
y = 3(-5)+2
y = -15+2
y = -13
the point (x,y) where the lines intersect is (-5,-13)
distance between (-5,-13) and (1,1)
d = sqrt[ (1 -(-5) )^2 + (1 - (-13) )^2 ]
d = sqrt[ (6)^2 + (14)^2 ]
d = sqrt[ 36 + 196 ]
d = sqrt[232]
d = 15.23154621
d = 15.2
CORRECT
in my view
i did
y = 3x + 2
slope > 3
distance > -1/3
distance has slope -1/3 passes through the point (1,1)
y - y1 = m(x - x1) < point slope formula
y - 1 = (-1/3)(x-1)
y - 1 = -1/3x + 1/3
y = -1/3x + 1 1/3
y = -1/3x + 4/3
y=-1/3x + 4/3
y = 3x + 2
find the intersection point
-1/3x + 4/3 = 3x + 2
-x + 4 = 9x + 6
-10x = 2
x = -5
y = 3(-5)+2
y = -15+2
y = -13
the point (x,y) where the lines intersect is (-5,-13)
distance between (-5,-13) and (1,1)
d = sqrt[ (1 -(-5) )^2 + (1 - (-13) )^2 ]
d = sqrt[ (6)^2 + (14)^2 ]
d = sqrt[ 36 + 196 ]
d = sqrt[232]
d = 15.23154621
d = 15.2
CORRECT