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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
 #1
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+2

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 :

 

 

 

cool cool cool

 Dec 21, 2020
 #2
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+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

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