Originally Posted by

**Jhevon** we still use the midpoint formula, but now we equate it to a result.

For T(-4,3) , S(-1,5)

Let the coordinates of $\displaystyle R$ be $\displaystyle (x_1,y_1)$

Since $\displaystyle \mbox {midpoint} \{R,T \} = S$ we have

$\displaystyle \left( \frac {x_1 - 4}{2}, \frac {y_1 + 3}{2} \right) = (-1,5)$

$\displaystyle \Rightarrow \frac {x_1 - 4}{2} = -1$ and $\displaystyle \frac {y_1 + 3}{2} = 5$

now solve for $\displaystyle x_1$ and $\displaystyle y_1$

the other question in this section is, of course, done the same way