The line y= -12/5x + 2 is exactly 8 units away from two other lines parallel to it. The distance in units between the y-intercepts of these two other lines is?
+129840 The y intercept of the given line = 2
The slope of this line = (-12/5)
A line perpendiculr to this one will have a slope of 5/12
And this line will pss through the point ( 8 ( 12/13) , 2 + 8(5/13) ) = ( 96/13, 66/13)
So the equation of a parrallel line is
y =(-12/5)(x - 96/13) + 66/13
y = (-12/5)x + 1152/65 + 66/13
y = (-12/5)x + 114/5
And because of symmetry.....the distance between the y intercepts = 2 (114/5 -2) = 2 ( 104/5) = 208/5
See the graph here :
Find equation of one of the other lines in y = mx + b form
From the given line -12/5 x + 2 the distance to the new line needs to be 8
PERPINDICULAR line is 5/12 x + b
distance along perpindicular line needs to be 8 from 0, 2
(x-0)^2 + (y-2)^2 = 8^2
use ratios to find the new point (you should draw a picture)
8/13 = x/12 x = 7.38462 from 0= 7.38462
8/13 = y/5 y = 3.0769 from y =2 = 5.0769 This is our point on the parallel line
that is 8 units from 0,2
y= mx+b
5.0769 = -12/5 (7.38462)+ b results in b = ~22.8 this is the y axis crossing which is 20.8 from the original line y intercept
the other parallel line is also 20.8 from the original intercept 20.8 + 20.8 =
41.6 units between the y-intercepts of the parallel lines
