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Zorluk: ZorReal Numbers, Number Line, and Absolute Value

On a real number line, point AA has coordinate −5-5 and point BB has coordinate 1111. Point CC has coordinate xx such that the distance between CC and the midpoint of line segment ABAB is equal to 13\frac{1}{3} of the distance between CC and point BB. What is the maximum possible value of xx?

Cevap: 5

Cevap

The maximum possible value of xx is 55.
The midpoint of A(−5)A(-5) and B(11)B(11) is M=3M = 3. Setting up the distance equation ∣x−3∣=13∣x−11∣|x - 3| = \frac{1}{3}|x - 11| leads to 3∣x−3∣=∣x−11∣3|x - 3| = |x - 11|. Evaluating the two cases 3(x−3)=x−113(x - 3) = x - 11 and 3(x−3)=−(x−11)3(x - 3) = -(x - 11) yields solutions x=−1x = -1 and x=5x = 5. The maximum possible value among these is 5.

Adım Adım Çözüm

1
Calculate the coordinate of the midpoint of line segment ABAB.
Midpoint coordinate M=3M = 3.
The midpoint of two coordinates aa and bb on a number line is given by a+b2=−5+112=3\frac{a + b}{2} = \frac{-5 + 11}{2} = 3.
2
Formulate the distance equation using absolute values.
3∣x−3∣=∣x−11∣3|x - 3| = |x - 11|.
The distance between xx and 33 is ∣x−3∣|x - 3| and the distance between xx and 1111 is ∣x−11∣|x - 11|. The given condition is ∣x−3∣=13∣x−11∣|x - 3| = \frac{1}{3}|x - 11|.
3
Solve the absolute value equation for all possible values of xx.
x=−1x = -1 and x=5x = 5.
Splitting 3(x−3)=x−113(x - 3) = x - 11 yields x=−1x = -1, and splitting 3(x−3)=−(x−11)3(x - 3) = -(x - 11) yields x=5x = 5.
4
Select the maximum value among all valid solutions.
55
Comparing x=−1x = -1 and x=5x = 5, the maximum value is 55.

Anahtar Kavram

Distance on a number line using absolute value and midpoint formula
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