Find the length of segment AB where A is (-1, 1) and B is (4, 6). Then find the midpoint of the segment joining (4, 1) and (16, -1).

Step-by-step solution

1. 1

   Keep the two formulas straight. They look similar but do opposite things:

   $d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\quad(\text{subtract, then square}),$
   $M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\quad(\text{add, then halve}).$

   Distance uses differences and is sign-proof; the midpoint uses sums, so a mishandled negative changes the answer.

2. 2

   Part (1): compute the differences.

   $x_2-x_1=4-(-1)=5,\qquad y_2-y_1=6-1=5.$

3. 3

   Part (1): apply the distance formula and simplify the radical.

   $AB=\sqrt{5^{2}+5^{2}}=\sqrt{50}=\sqrt{25\cdot 2}=5\sqrt{2}\approx 7.071.$

   Equal differences mean the segment makes a $45^\circ$ angle with the axes, which is why the answer is a clean multiple of $\sqrt2$.

4. 4

   Part (2): average the x-coordinates.

   $\frac{4+16}{2}=\frac{20}{2}=10.$

5. 5

   Part (2): average the y-coordinates, watching the negative.

   $\frac{1+(-1)}{2}=\frac{0}{2}=0.$

   So the midpoint is $M(10,0)$ — on the $x$-axis, which makes sense since the two points sit symmetrically one unit above and one unit below it.

6. 6

   Check the midpoint by distance. From $(4,1)$ to $(10,0)$ is $\sqrt{36+1}=\sqrt{37}$, and from $(10,0)$ to $(16,-1)$ is $\sqrt{36+1}=\sqrt{37}$ as well. Equal distances confirm $M$ really is the midpoint.