Question

Points P and Q belong to segment AB . If AB = a, AP = 2PQ = 2QB, find the distance: midpoints between AP QB

Answer 1

Answer: The distace between midpoints of AP and QB is $\frac{a}{8}$.

Step-by-step explanation: Points P and Q are between points A and B and the segment AB measures a, then:

AP + PQ + QB = a

According to the question, AP = 2 PQ = 2QB, so:

PQ = $\frac{AP}{2}$

QB = $\frac{AP}{2}$

Substituing:

AP + 2*($\frac{AP}{2}$) = a

2AP = a

AP = $\frac{a}{2}$

Since the distance is between midpoints of AP and QB:

2QB = AP

QB = $\frac{AP}{2}$

QB = $\frac{a}{2}*\frac{1}{2}$

QB = $\frac{a}{4}$

MIdpoint is the point that divides the segment in half, so:

Midpoint of AP:

$\frac{AP}{2} = \frac{a}{2}*\frac{1}{2}$

$\frac{AP}{2} = \frac{a}{4}$

Midpoint of QB:

$\frac{QB}{2} = \frac{a}{4}*\frac{1}{2}$

$\frac{QB}{2} = \frac{a}{8}$

The distance is:

d = $\frac{a}{4} - \frac{a}{8}$

d = $\frac{a}{8}$