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# lines

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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?

Dec 21, 2020

### 2+0 Answers

#1
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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 :    Dec 21, 2020
#2
+1

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

Dec 21, 2020