T. Pitagora twice => new street = 2

= 8.94 miles;
135*8.94 = 1206.9$;
By definition, if two lines share the same gradient, they are said to be parallel. So, we know for this equation, it must have a gradient of 1/2.
Now, since the point (-6, 4) passes through the line, we know it must satisfy the equation. Since we have a gradient/slope and a point, we can use the point-gradient form:

, where

represents the points being passed through.



The perimeter without the diagonal is 4*120 = 480
That covers the outside of the field
The diagonal is 120 *

= 169.7 yd
480 + 169.7 = 649.7 yd total of fencing