Question

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use the figure for exercises 1–4. \\(\overline{ab}\\) is a midsegment of \\(\triangle rst\\).

1. what is the slope of midsegment \\(\overline{ab}\\) and the slope of side \\(\overline{st}\\)?

1. what can you conclude about \\(\overline{ab}\\) and \\(\overline{st}\\)?

1. find \\(ab\\) and \\(st\\).

1. compare the lengths of \\(\overline{ab}\\) and \\(\overline{st}\\).

use \\(\triangle mnp\\) for exercises 5–7.

Explanation:

Question 1

Step1: Find slope of AB

Points \( A(1, 0) \) and \( B(3, -2) \). Slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
\( m_{AB} = \frac{-2 - 0}{3 - 1} = \frac{-2}{2} = -1 \).

Step2: Find slope of ST

Points \( S(2, 3) \) and \( T(6, -1) \).
\( m_{ST} = \frac{-1 - 3}{6 - 2} = \frac{-4}{4} = -1 \).

Since slopes of \( \overline{AB} \) and \( \overline{ST} \) are equal (\( -1 \)), by the definition of parallel lines (same slope), \( \overline{AB} \parallel \overline{ST} \).

Step1: Length of AB

Points \( A(1, 0) \), \( B(3, -2) \). Distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
\( AB = \sqrt{(3 - 1)^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \).

Step2: Length of ST

Points \( S(2, 3) \), \( T(6, -1) \).
\( ST = \sqrt{(6 - 2)^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \).

Answer:

Slope of \( \overline{AB} \) is \(-1\), slope of \( \overline{ST} \) is \(-1\).

Question 2