1. Standardized Test Practice
EXAMPLE 2
Standardized Test Practice
SOLUTION
By counting, PQ = 4 and QR = 3. Use the Distance Formula to find PR. d = y 2 – 1 ( ) x +

2. Standardized Test Practice
EXAMPLE 2
Standardized Test Practice = + 1 – 4 ( ) 2
– 1
(– 5
) ) PR = 4 2 +
(– 3 ) = 25 5 =
By the SSS Congruence Postulate, any triangle with side lengths 3, 4, and 5 will be congruent to
PQR.
The distance from (–1, 1) to (–1, 5) is 4. The distance from (–1, 5) to (–4, 5) is 3. The distance from (– 1, 1) to (–4, 5) is = 4 2 +
(– 3 5
(–4) – (–1) ( )
5 – 1) 25
The correct answer is A.
ANSWER

3. GUIDED PRACTICE
for Example 2 4.
has vertices J(–3, –2), K(0, –2), and L(–3, –8).
RST has vertices R(10, 0), S(10, – 3), and T(4, 0).
Graph the triangles in the same coordinate plane and show that they are congruent.
JKL
ANSWER
KJ = SR = 3.
JL = RT = 6.
LK = TS = 3 5.