Segments, Length, and Distance: Practice

Add the adjacent parts and set them equal to the whole.

$(2x+3)+(x+5)=29$

Combine terms to get $3x+8=29$.

Subtract $8$ and divide by $3$, giving $x=7$.

The parts are $17$ and $12$, both positive, and add to $29$.

The midpoint makes the half lengths equal: $4x-1=2x+9$, so $2x=10$ and $x=5$. Each half has length $19$.

$PQ=19+19=38$

The midpoint is $11$ units to the right of $-15$. Continue the same distance to the right.

$-4+11=7$

The midpoint is between the endpoints.

A given distance allows a point in either direction on the line.

$-6-9=-15$

$-6+9=3$

Both coordinates are $9$ units from $-6$.

The middle point is $B$, so $AB$ and $BC$ are the two adjacent parts. Their sum is the whole length $AC$.

A bisector must pass through the midpoint. Here the two parts have unequal lengths, $6$ and $11$, so $N$ is not the midpoint.

Congruence gives $5x+4=3x+16$, so $2x=12$ and $x=6$. Substitute into either length.

$AB=5(6)+4=34\text{ cm}$

The other expression gives $3(6)+16=34$ centimeters too.

Subtract the readings in centimeters, then convert.

$6.4-1.6=4.8\text{ cm}$

$4.8\times10=48\text{ mm}$

Convert $3.5$ centimeters to $35$ millimeters before adding the adjacent lengths.

$UW=35+28=63\text{ mm}$

A physical reading has limited precision. Recording nearest millimeters supports an approximate value of $52$ millimeters, not an exact claim with arbitrarily many decimal places.

The coordinate $1$ lies between the endpoints. The adjacent gaps are $9$ and $11$.

$9+11=20$

This matches the whole distance $12-(-8)=20$.

A midpoint must lie on the segment and divide it into equal lengths. Both conditions are stated for $M$.

Although $AN=NB$, point $N$ lacks the required location on the segment, so it is not the midpoint.