★ AIMS Think+ · Q11

Given:

AP = PQ = QB along AB.

To Find:

P and Q.

Property:

Apply the midpoint relationship twice, coordinate by coordinate.

Solution:

For either coordinate, 2p = a + q and 2q = p + b.

Substitute q = 2p − a into the second equation: 4p − 2a = p + b, hence p = $\frac{2a+b}{3}$.

Similarly q = $\frac{a+2b}{3}$.

P = ($\frac{2\times 4+16}{3}$, $\frac{2\times 7-2}{3}$) = $\left(8,4\right)$.

Q = ($\frac{4+2\times 16}{3}$, $\frac{7+2\times \left(-2\right)}{3}$) = $\left(12,1\right)$.

Successive coordinate changes are all $\left(4,-3\right)$, so each small segment has length 5.

Results:

P$\left(8,4\right)$, Q$\left(12,1\right)$.