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.
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
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 !