Question

Point C ∈ AB, AB = 30m. Point C is 1.5 times farther from point A than from point B. Find AC and CB.

Answer 1

We are given that

point C lies on AB

Let's assume AC=x

we know that

CB=AB-AC

$CB=30-x$

Point C is 1.5 times farther from point A than from point B

so, we get

$\frac{AC}{CB} =1.5$

now, we can plug values

$\frac{x}{30-x} =1.5$

now, we can solve for x

$x=1.5\left(30-x\right)$

$x=18$

so, we get

$AC=18$

now, we can find CB

$CB=30-18$

$CB=12$

so, we get

$AC=18$

$CB=12$...............Answer