Question

Difficulty: HardReal Numbers, Number Line, and Absolute Value

On the real number line, points PP, QQ, and RR have coordinates xx, yy, and zz, respectively, such that ∣x−2∣=5|x - 2| = 5, ∣y+4∣=3|y + 4| = 3, and zz is the midpoint of segment PQPQ. If x<yx < y, what is the value of zz?

Answer: -2

Answer

The coordinate of point RR (the value of zz) is −2-2.
Solving the two absolute value equations yields x∈{−3,7}x \in \{-3, 7\} and y∈{−7,−1}y \in \{-7, -1\}. Testing the given constraint x<yx < y across all four possible ordered pairs reveals that only x=−3x = -3 and y=−1y = -1 satisfy the condition, since −3<−1-3 < -1 is true while all other pairs fail. The midpoint of points with coordinates −3-3 and −1-1 is −3+(−1)2=−2\frac{-3 + (-1)}{2} = -2.

Step-by-Step Solution

1
Solve the absolute value equation ∣x−2∣=5|x - 2| = 5.
x=7x = 7 or x=−3x = -3
The equation ∣x−2∣=5|x - 2| = 5 splits into x−2=5  ⟹  x=7x - 2 = 5 \implies x = 7 and x−2=−5  ⟹  x=−3x - 2 = -5 \implies x = -3.
2
Solve the absolute value equation ∣y+4∣=3|y + 4| = 3.
y=−1y = -1 or y=−7y = -7
The equation ∣y+4∣=3|y + 4| = 3 splits into y+4=3  ⟹  y=−1y + 4 = 3 \implies y = -1 and y+4=−3  ⟹  y=−7y + 4 = -3 \implies y = -7.
3
Evaluate all candidate pairs (x,y)(x, y) under the inequality constraint x<yx < y.
The only valid pair is x=−3x = -3 and y=−1y = -1.
Comparing all four combinations: 7<−17 < -1 (false), 7<−77 < -7 (false), −3<−7-3 < -7 (false), and −3<−1-3 < -1 (true).
4
Calculate the midpoint zz of the segment PQPQ.
z=−2z = -2
The midpoint of coordinates −3-3 and −1-1 on a number line is given by their average: −3+(−1)2=−2\frac{-3 + (-1)}{2} = -2.

Key Concept

Absolute Value Equations and Midpoint on the Real Number Line
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