what is the distance from point (-6,4) to the line y=1/3x-4

Question

what is the distance from point (-6,4) to the line y=1/3x-4

Answer 1

write line in ax + by + c = 0 format:

3y = x - 12

x - 3y - 12 = 0

a = 1, b = -3, c = -12

point (x0,y0) = (-6,4)

distance = (|ax0 + By0 + c|)/sqrt(a^2 + b^2)

(|1 * -6 + -3 * 4 + -12|)/ sqrt(1 * 1 + -3 * -3)

30/sqrt(10) = 3 sqrt(10) = 9.49

Answer 2

distance from point (-6,4) to the line y=1/3x-4

=|(1/3)(-6)-4-4|/(1+1/9)^(1/2)

=|mx₀+b-y₀|/(1+m²)^(1/2) ,,,,provable with analytical geometry and similar triangles

m=(1/3),,x₀=-6,,y₀=4

corrected

distance=(30/√10)=3√10

Answer 3

y=-3x +b

4=-3(-6)+b

b=-14

y=-3x -14 is the perpendicular line thru the point

find where the lines intersect

y=-3x-14= x/3 -4

10x/3= -10

x=-3

y=-5

from (-3,-5) to (-6.4) is sqr(90)=3sqr10

= about 9.49 = distance from point to the line

try plotting the point & graph the line to check that the solution looks reasonable