Point M(2, 2) is the midpoint of segment RS. If R(-1, 4) is one endpoint, what are the coordinates of endpoint S?

Step-by-step solution

1. 1

   Reverse the midpoint formula. From $x_M = \dfrac{x_1+x_2}{2}$, multiplying by $2$ and rearranging gives

   $x_2 = 2x_M - x_1, \qquad y_2 = 2y_M - y_1$

   So double the midpoint and subtract the known endpoint. Averaging $M$ with $R$ instead — a very common reflex — gives the midpoint of $RM$, not $S$.

2. 2

   Apply it to the $x$-coordinate.

   $x_S = 2(2) - (-1) = 4 + 1 = 5$

   The double negative is where marks are lost: subtracting $-1$ adds $1$.

3. 3

   Apply it to the $y$-coordinate.

   $y_S = 2(2) - 4 = 4 - 4 = 0$

4. 4

   State the endpoint.

   $S = (5,\ 0)$

5. 5

   Check with the step-vector view. Going from $R(-1,4)$ to $M(2,2)$ moves $\Delta x = +3$, $\Delta y = -2$. Repeating that same step from $M$ lands at $(2+3,\ 2-2) = (5,0)$ ✓ — which is exactly what "midpoint" means: the same step twice.

6. 6

   Verify with the forward formula. Midpoint of $R(-1,4)$ and $S(5,0)$: $\left(\dfrac{-1+5}{2},\ \dfrac{4+0}{2}\right) = (2,\,2)$ ✓, recovering $M$. Distractors $(1,3)$ and $(3,0)$ fail this test.