Question

mixed exercises
find the coordinates of the missing endpoint if p is the mid - point of nq.

1. n(2,0), p(5,2)
2. n(5,4), p(6,3)
3. q(3,9), p(-1,5)

Explanation:

Step1: Recall mid - point formula

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $P(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$, where $P$ is the mid - point. Let $N(x_1,y_1)$ and $Q(x_2,y_2)$, and $P$ be the mid - point. Then $x_P=\frac{x_1 + x_2}{2}$ and $y_P=\frac{y_1 + y_2}{2}$.

Step2: Solve for 49

Let $N(x_1,y_1)=(2,0)$ and $P(x_P,y_P)=(5,2)$.
For the $x$ - coordinate: $x_P=\frac{x_1 + x_2}{2}$, so $5=\frac{2 + x_2}{2}$, then $10 = 2+x_2$, and $x_2=8$.
For the $y$ - coordinate: $y_P=\frac{y_1 + y_2}{2}$, so $2=\frac{0 + y_2}{2}$, then $y_2 = 4$. The coordinates of $Q$ are $(8,4)$.

Step3: Solve for 50

Let $N(x_1,y_1)=(5,4)$ and $P(x_P,y_P)=(6,3)$.
For the $x$ - coordinate: $x_P=\frac{x_1 + x_2}{2}$, so $6=\frac{5 + x_2}{2}$, then $12=5 + x_2$, and $x_2 = 7$.
For the $y$ - coordinate: $y_P=\frac{y_1 + y_2}{2}$, so $3=\frac{4 + y_2}{2}$, then $6 = 4+y_2$, and $y_2=2$. The coordinates of $Q$ are $(7,2)$.

Step4: Solve for 51

Let $Q(x_1,y_1)=(3,9)$ and $P(x_P,y_P)=(-1,5)$.
For the $x$ - coordinate: $x_P=\frac{x_1 + x_2}{2}$, so $-1=\frac{3 + x_2}{2}$, then $-2=3 + x_2$, and $x_2=-5$.
For the $y$ - coordinate: $y_P=\frac{y_1 + y_2}{2}$, so $5=\frac{9 + y_2}{2}$, then $10 = 9+y_2$, and $y_2 = 1$. The coordinates of $N$ are $(-5,1)$.

Answer:

1. $(8,4)$
2. $(7,2)$
3. $(-5,1)$