Question

in exercises 19 - 22, the mid - point m and one endpoint of gh are given. find the coordinates of the other endpoint. (see example 3.) 19. g(5, - 6) and m(4, 3) 20. h(- 3, 7) and m(- 2, 5) 21. h(- 2, 9) and m(8, 0) 22. g(- 4, 1) and m(-\frac{13}{2}, - 6)

Explanation:

Step1: Recall mid - point formula

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.

Step2: Solve for unknown coordinates

Let one endpoint be $(x_1,y_1)$ and mid - point be $(x_m,y_m)$. Then $x_2 = 2x_m−x_1$ and $y_2 = 2y_m−y_1$.

19.
Let $G(5,-6)$ be $(x_1,y_1)$ and $M(4,3)$ be $(x_m,y_m)$.
$x_2=2\times4 - 5=3$
$y_2=2\times3+6 = 12$
The other endpoint is $(3,12)$.

20.
Let $H(-3,7)$ be $(x_1,y_1)$ and $M(-2,5)$ be $(x_m,y_m)$.
$x_2=2\times(-2)+3=-1$
$y_2=2\times5 - 7 = 3$
The other endpoint is $(-1,3)$.

21.
Let $H(-2,9)$ be $(x_1,y_1)$ and $M(8,0)$ be $(x_m,y_m)$.
$x_2=2\times8 + 2=18$
$y_2=2\times0 - 9=-9$
The other endpoint is $(18,-9)$.

22.
Let $G(-4,1)$ be $(x_1,y_1)$ and $M(-\frac{13}{2},-6)$ be $(x_m,y_m)$.
$x_2=2\times(-\frac{13}{2})+4=-9$
$y_2=2\times(-6)-1=-13$
The other endpoint is $(-9,-13)$.

Answer:

1. $(3,12)$
2. $(-1,3)$
3. $(18,-9)$
4. $(-9,-13)$