Point A is two-thirds of the way along the segment from (1, 10) to (-8, 4). Point B is the midpoint of the segment joining C(0, -7) and D(6, -11). Find the distance AB.

Step-by-step solution

1. 1

   Set up the section formula for A. Moving a fraction $t$ of the way from $P$ to $Q$ lands at $P+t(Q-P)$. Here $t=\frac23$, $P=(1,10)$ and $Q=(-8,4)$, so the displacement is $Q-P=(-9,-6)$.

2. 2

   Compute A. $A=(1,10)+\frac23(-9,-6)=(1-6,\ 10-4)=(-5,6)$. Both coordinates come out whole because $-9$ and $-6$ are divisible by $3$.

3. 3

   Compute B with the midpoint formula. The midpoint averages the coordinates: $B=\left(\frac{0+6}{2},\frac{-7+(-11)}{2}\right)=(3,-9)$.

4. 4

   Apply the distance formula. $AB=\sqrt{(3-(-5))^2+(-9-6)^2}=\sqrt{8^2+(-15)^2}=\sqrt{64+225}=\sqrt{289}$.

5. 5

   Simplify the radical. $\sqrt{289}=17$, so $AB=17$. The legs $8$ and $15$ with hypotenuse $17$ form a well-known Pythagorean triple, which is a good sign the arithmetic is right.

6. 6

   Check that A really is two-thirds along. The distance from $(1,10)$ to $A(-5,6)$ is $\sqrt{36+16}=\sqrt{52}\approx 7.211$, and the full segment $(1,10)$ to $(-8,4)$ has length $\sqrt{81+36}=\sqrt{117}\approx 10.817$. The ratio $7.211/10.817=0.6667$, exactly two-thirds.