Question

points p, q, and r are collinear. point q is between p and r, pq = 7z + 4, pr = 14z − 14, and qr = 6z + 11. find pq, qr, and pr. pq = qr = pr =

Explanation:

Step1: Use segment addition postulate

Since Q is between P and R, \( PR = PQ + QR \). Substitute the given expressions: \( 14z - 14=(7z + 4)+(6z + 11) \)

Step2: Simplify and solve for z

Simplify the right - hand side: \( 14z - 14 = 13z+15 \). Subtract \( 13z \) from both sides: \( 14z-13z - 14=13z - 13z+15 \), which gives \( z - 14 = 15 \). Add 14 to both sides: \( z=15 + 14=29 \)

Step3: Find PQ

Substitute \( z = 29 \) into \( PQ=7z + 4 \): \( PQ=7\times29+4=203 + 4=207 \)

Step4: Find QR

Substitute \( z = 29 \) into \( QR = 6z+11 \): \( QR=6\times29+11=174 + 11=185 \)

Step5: Find PR

Substitute \( z = 29 \) into \( PR=14z - 14 \): \( PR=14\times29-14=406-14 = 392 \) (We can also verify using \( PR=PQ + QR=207 + 185 = 392 \))

Answer:

\( PQ = 207 \), \( QR = 185 \), \( PR = 392 \)