Question

1. find the length of each segment in the path shown.

what is the total distance?
example: e(2, -1) and f(6, 0). the distance d from point e to point f is
calculated with the distance formula.
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$= \sqrt{(6 - 2)^2 + (0 - (-1))^2}$
$= \sqrt{(4)^2 + (1)^2}$
$= \sqrt{16 + 1}$
$= \sqrt{17} \approx 4.123$

Explanation:

Step1: Find coordinates of points

Assume \(A(-4,8)\), \(B(-4,6)\), \(C(-2,6)\), \(D(1,3)\), \(E(1,-1)\), \(F(6,0)\).

Step2: Calculate length of \(AB\)

Using vertical distance formula \(d = |y_2 - y_1|\).
\(AB=\vert8 - 6\vert=2\)

Step3: Calculate length of \(BC\)

Using horizontal distance formula \(d = |x_2 - x_1|\).
\(BC=\vert- 2-(-4)\vert = 2\)

Step4: Calculate length of \(CD\)

Using distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(x_1=-2,y_1 = 6,x_2=1,y_2 = 3\)
\(CD=\sqrt{(1-(-2))^2+(3 - 6)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\approx4.243\)

Step5: Calculate length of \(DE\)

Using vertical distance formula \(d = |y_2 - y_1|\)
\(DE=\vert-1-3\vert = 4\)

Step6: Calculate total distance

Total distance \(=AB + BC+CD+DE+EF\)
\(=2 + 2+4.243+4 + 4.123\)
\(=16.366\)

Answer:

\(AB = 2\), \(BC = 2\), \(CD\approx4.243\), \(DE = 4\), \(EF\approx4.123\), total distance \(\approx16.366\)