Question

diego and jayson are at the mall. diego is at the bucks entrance (-1, 8) and jayson is at game stop. what would be the midpoint for diego and jayson to meet?

Explanation:

Step1: Recall mid - point formula

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here, assume the coordinates of Diego are $(x_1,y_1)=(-1,8)$ and let the coordinates of Jayson be $(x_2,y_2)$. Since no coordinates for Jayson are given other than the context of being at Game Stop, assume for simplicity (if we consider this as a 2 - D coordinate problem) that we need to find the mid - point with respect to the given point of Diego. Let's assume Jayson's coordinates are $(x_2,y_2)=(x,y)$. The mid - point $M$ has coordinates $M=(\frac{-1 + x}{2},\frac{8 + y}{2})$. If we assume Jayson's coordinates are not given and we just want to state the general mid - point formula application, or if we assume Jayson's coordinates are also known values. Let's assume Jayson is at $(x_2,y_2)=(x_2,y_2)$.
The mid - point $M$ of the two points $(-1,8)$ and $(x_2,y_2)$ is calculated as follows.
For the x - coordinate of the mid - point:
Let $x_m=\frac{-1+x_2}{2}$.
For the y - coordinate of the mid - point:
Let $y_m=\frac{8 + y_2}{2}$.

Step2: Assume a common case

If we assume Jayson's coordinates are, say, $(x_2,y_2)=(3,4)$ (random example for illustration).
For the x - coordinate of the mid - point:
$x_m=\frac{-1 + 3}{2}=\frac{2}{2}=1$.
For the y - coordinate of the mid - point:
$y_m=\frac{8+4}{2}=\frac{12}{2}=6$.
The mid - point is $(1,6)$. In general, if Jayson's coordinates are $(x_2,y_2)$, the mid - point is $(\frac{-1 + x_2}{2},\frac{8 + y_2}{2})$.

Answer:

If Jayson's coordinates are $(x_2,y_2)$, the mid - point is $(\frac{-1 + x_2}{2},\frac{8 + y_2}{2})$