Using Midpoint and Distance Formulas

We know that $\overline{A B}$ and $\overline{B C}$ have the same length because they have matching tick-marks. Therefore, the line that goes through the midpoint $B$ is the segment bisector. This is line $n .$

Since

A B

and

B C

have equal length we can set there expressions equal to each other and solve for

x .

4 x + 8 = 10 x + 12

SubEqn

LHS - 4 x = RHS - 4 x

8 = 6 x + 12

SubEqn

LHS - 12 = RHS - 12

- 4 = 6 x

DivEqn

LHS / 6 = RHS / 6

\frac{- 4}{6} = x

RearrangeEqnRearrange equation

x = \frac{- 4}{6}

MoveNegNumToFracPut minus sign in front of fraction

x = - \frac{4}{6}

ReduceFrac

\frac{a}{b} = \frac{a / 2}{b / 2}

x = - \frac{2}{3}

The length of

\overline{A C}

is given by

4 x + 8 + 10 x + 12,

the sum of

A B

and

B C .

We can simplify this expression and then substitute

- \frac{2}{3}

for

x

to find

A C .

A C = 4 x + 8 + 10 x + 12

AddTermsAdd terms

A C = 14 x + 20

Substitute

x = - \frac{2}{3}

A C = 14 (- \frac{2}{3}) + 20

MultPosNeg

a (- b) = - a \cdot b

A C = - 14 \cdot \frac{2}{3} + 20

MultiplyMultiply

A C = - \frac{2 8}{3} + 20

NumberToFrac

a = \frac{3 \cdot a}{3}

A C = - \frac{2 8}{3} + \frac{6 0}{3}

AddTermsAdd terms

A C = \frac{3 2}{3}

The length of

\overline{A C}

is

\frac{3 2}{3}

units.

Using Midpoint and Distance Formulas 1.5 - Solution