On the real number line, points P, Q, and R have coordinates x, y, and z, respectively, su… — GRE General Test

On the real number line, points $P$ , $Q$ , and $R$ have coordinates $x$ , $y$ , and $z$ , respectively, such that $|x - 2| = 5$ , $|y + 4| = 3$ , and $z$ is the midpoint of segment $PQ$ . If $x < y$ , what is the value of $z$ ?

Answer: -2

Answer

The coordinate of point

R

(the value of

z

) is

-2

Solving the two absolute value equations yields

x \in \{-3, 7\}

and

y \in \{-7, -1\}

. Testing the given constraint

x < y

across all four possible ordered pairs reveals that only

x = -3

and

y = -1

satisfy the condition, since

-3 < -1

is true while all other pairs fail. The midpoint of points with coordinates

-3

and

-1

\frac{-3 + (-1)}{2} = -2

Step-by-Step Solution

Solve the absolute value equation

|x - 2| = 5

x = 7

x = -3

The equation

|x - 2| = 5

splits into

x - 2 = 5 \implies x = 7

and

x - 2 = -5 \implies x = -3

Solve the absolute value equation

|y + 4| = 3

y = -1

y = -7

The equation

|y + 4| = 3

splits into

y + 4 = 3 \implies y = -1

and

y + 4 = -3 \implies y = -7

Evaluate all candidate pairs

(x, y)

under the inequality constraint

x < y

The only valid pair is

x = -3

and

y = -1

Comparing all four combinations:

7 < -1

(false),

7 < -7

(false),

-3 < -7

(false), and

-3 < -1

(true).

Calculate the midpoint

z

of the segment

PQ

z = -2

The midpoint of coordinates

-3

and

-1

on a number line is given by their average:

\frac{-3 + (-1)}{2} = -2

Key Concept

Absolute Value Equations and Midpoint on the Real Number Line

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