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Explanation:
I'll define these point labels
The locations of the points are
Use the distance formula to find the distance from B to C
[tex]B = (x_1,y_1) = (-8,1) \text{ and } C = (x_2,y_2) = (-6,-3)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-8-(-6))^2 + (1-(-3))^2}\\\\d = \sqrt{(-8+6)^2 + (1+3)^2}\\\\d = \sqrt{(-2)^2 + (4)^2}\\\\d = \sqrt{4 + 16}\\\\d = \sqrt{20} \ \text{ ... exact distance}\\\\d \approx 4.47214 \ \ \text{... approximate distance}\\\\[/tex]
Segment BC is roughly 4.47214 units long.
Following similar steps, you should find that segment CF is approximately 10.04988 units long.
If Blake doesn't take the shortcut, then he travels approximately BC+CF = 4.47214+10.04988 = 14.52202 units. This is the path from B to C to F in that order.
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Use the distance formula again to find the distance from B to F. This distance is about 12.36932 units. He travels this amount if he takes the shortcut.
Subtract this and the previous result we got
14.52202 - 12.36932 = 2.1527
That rounds to 2.2
This is the amount of distance he doesn't have to travel when he takes the shortcut.
In other words, the track is roughly 2.2 units shorter when taking the shortcut.
Side note: Replace "units" with whatever units you're working with (eg: feet or meters).