We are given that
point C lies on AB
Let's assume AC=x
we know that
CB=AB-AC
[tex]CB=30-x[/tex]
Point C is 1.5 times farther from point A than from point B
so, we get
[tex]\frac{AC}{CB} =1.5[/tex]
now, we can plug values
[tex]\frac{x}{30-x} =1.5[/tex]
now, we can solve for x
[tex]x=1.5\left(30-x\right)[/tex]
[tex]x=18[/tex]
so, we get
[tex]AC=18[/tex]
now, we can find CB
[tex]CB=30-18[/tex]
[tex]CB=12[/tex]
so, we get
[tex]AC=18[/tex]
[tex]CB=12[/tex]...............Answer
