Problem 68

Solution 1

Let the coordinates of $R$ be $(a,b)$.

Since $Q$ is the midpoint of $P$ and $R$, then the difference between the $x$-coordinates of $Q$ and $P$ equals the difference between the $x$-coordinates of $R$ and $Q$.

In other words, $a-4 = 4-1$ and so $a=7$.

Similarly, $b - 7 = 7 - 3$ and so $b = 11$.

Thus, the coordinates of $R$ are $(7,11)$.

Solution 2

Let the coordinates of $R$ be $(a,b)$.

The midpoint of $P(1,3)$ and $R(a,b)$ has coordinates $\left(\frac{1}{2}(1+a),\frac{1}{2}(3+b)\right)$.

Since $Q(4,7)$ is the midpoint of $PR$, then $4 = \frac{1}{2}(1+a)$ (which gives $7 = 1+a$ or $a=7$) and $14 = \frac{1}{2}(3+b)$ (which gives $14=3+b$ or $b=11$).

Therefore, the coordinates of $R$ are $(7,11)$.