1. EXAMPLE 1 Standardized Test Practice SOLUTION Let ( x 1, y 1 ) = ( –3, 5) and ( x 2, y 2 ) = ( 4, – 1 ). = (4 – (–3)) 2 + (– 1 – 5) 2 = 49 + 36 = 85 ( x 2 – x 1 ) 2 + ( y 2 – y 1 ) 2 = d ANSWER The correct answer is C.

2. EXAMPLE 2 Classify a triangle using the distance formula ANSWER Because BC = AC, ∆ABC is isosceles. Classify ∆ABC as scalene, isosceles, or equilateral. AB=(7 – 4) 2 + (3 – 6) 2 = 18= 3 2BC=(2 – 7) 2 + (1 – 3) 2 = 29AC=(2 – 4) 2 + (1 – 6) 2 = 29

3. GUIDED PRACTICE for Examples 1 and 2 SOLUTION Let ( x 1, y 1 ) = ( 3, – 3) and ( x 2, y 2 ) = (–1, 5). 1. What is the distance between (3, – 3) and (– 1, 5) ? = (–1 – 3) 2 + ( 5 – (–3)) 2 = 16 + 64 ( x 2 – x 1 ) 2 + ( y 2 – y 1 ) 2 = d = 80 4 5 ANSWER The distance between (3, –3) and (–1, 5) is

4. GUIDED PRACTICE 2. The vertices of a triangle are R(– 1, 3), S(5, 2), and T(3, 6). Classify ∆RST as scalene, isosceles, or equilateral. R – 1,3 S 5, 2 T 3, 6 for Examples 1 and 2

5. GUIDED PRACTICE SOLUTION for Examples 1 and 2 ST=(3 – 5) 2 + (6 – 2) 2 = 20TR=(–1 –(–3) 2 + (3 – 6) 2 = 25 = 5 RS=(5 – (–1) 2 + (2 – 3) 2 = 36 = 6 ANSWERBecause RS ≠ ST ≠ TR, so RST is an scalene triangle.