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The line y= -12/5x + 2 is exactly 3 units away from two other lines parallel to it. The distance in units between the y-intercepts of these two other lines is.

 Jan 14, 2018
 #1
avatar+111329 
+2

 

 

To find these points..... first  find  the x  intersection of the graphs of

 

y  = (5/12)x  +  2     and    x^2  +  (y -2)^2  = 9

 

Manipulating the first....we have that  y  -  2  =  (5/12)x

 

Subbing this into the second, we have that

 

x^2  +  (5/12* x)^2  =  9      simplify

 

x^2  +  25x^2/ 144  = 9 

 

x^2  +  (25/144)x^2  = 9

 

169/144x^2  =  9

 

x^2  =  9*144/ 169       take the positive and negative roots

 

x  = ±√  [ 9 * 144 / 169 ]

 

x =  ± 36/13

 

So..using   y   =  (5/12)x  + 2...the y intersection points are

 

y  =  (5/12)(36/13)  +  2  =  (180/156) + 2  =  41/13

 

And 

 

y  =   (5/12)(-36/13)  +  2  =  (-180/156)  + 2  =  11/13

 

So....the points of intersection are

 

(36/13, 41/13)   and  (  -36/13, 11/13)

 

So  using these   points and the slope  (-12/5)   we have the  equations

 

 

y  =  (-12/5)( x - 36/13)  +41/13

So...when   x  = 0, the y intercept of this line  is  (-12/5)(-36/13)  + 41/13  = 49/5  = 9.8

 

And

 

y  =  (-12/5)( x +  36/13)  +  11/13

And.....when x  = 0, the y intercept of this line  is (-12/5)(36/13)  + 11/13  = -29/5  = -5.8 

 

Here's  a graph : 

https://www.desmos.com/calculator/ctru5i68ra

 

 

cool cool cool

 Jan 15, 2018
 #2
avatar+23646 
0

I was looking at this question the other day and thinking of an easy way to solve it....then I saw CPHIL's answer which was different than my thinking, but I wil use his graph diagram rather than uploading my hand drawings.....

The short thick black line is = 3  (given in the question)   the slope of the blue line is - 12/5  (given)....  Now let's rotate the triangle so the blue line is verticle ..like this:

Now the thin black line (hypotenuse, h) that we are looking for has a slope of 12/5 

We can find the length of the blue line, b, by  TAN = opp/adj = 12/5 = b/3     so b = 36/5   

Now, use Pythag theorom to find the hypotenuse (the distance on the y axis betweeen the two parallel lines)

h^2= (36/5)^2  + 3^2    results in h= 7.8

we need TWO h's to find the distance between the TWO lines  parallel to the given line:     2 x 7.8 = 15.6 

 

Hope this isn't too confusing !

 Jan 15, 2018
 #3
avatar+111329 
0

My  answer was the same as EPs   =  9.8  +  l - 5.8 l   =  15.6

 

His solution is probably  more elegant, though  !!!

 

 

cool cool cool

 Jan 15, 2018

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