Fine the distance of point (8,-2) from the line 6x - 8y +6 = 0

Fine the distance of point (8,-2) from the line 6x - 8y +6 = 0

Any help would be appreciated x

Fine the distance of point (8,-2) from the line 6x - 8y +6 = 0

first, the slope of the line
6x-8y+6=0
6x+6=8y

6x/8+6/8=y
y=(3/4)x+[does not matter]

slope=3/4

distance from the point will be on a line perpendicular from the line
perpendicular is the negative inverse
m= -4/3

so you start with
y= -4x/3

to pass through the point
SUBTRACT the values from their relative variable
(8,-2)
(y-(-2))= -4(x-8)/3
y+2= -4x/3+32/3

y= -4x/3+26/3

oops, you wanted the distance
so, now we have to find out where the lines intersect
y= -4x/3+26/3
3y= -4x+26
4x+3y=26

the two lines are
4x+3y=26
6x-8y= -6

you have +/-y
multiply the first equation by 8
and the second by 3
then add
32x+24y=208
18x-24y= -18
50x=190
x=190/50=19/5

use the smaller x coefficient
4x+3y=26
760/50+3y=26
76/5+3y=130/5
3y=54/5
y=18/5

the lines intersect at(19/5,18/5)

distance from (8,-2) to (19/5,18/5)
[if I am doing this right....
your teacher hates you] LOL

distance formula
c^2=a^2+b^2
c^2=(8-19/5)^2+(-2-18/5)^2
c^2=(40/5-19/5)^2+(-10/5-18/5)^2
c^2=(21/5)^2+(-28/5)^2
c^2=(441/25)+(784/25)
c^2=1225/25
c=35/5

the distance is 7

Just plug in the 8 and -2 6(8) -8(-2)+6. 48+16+6= 70