Question

the midpoint m and one endpoint of cd are given. find the coordinates of the second endpoint.

1. c(5, -6) and m(4,3)
2. m(-3,7) and d(-2,5)
3. m(-2,9) and d(8,0)

Explanation:

Step1: Recall mid - point formula

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Let the known endpoint be $(x_1,y_1)$ and the unknown endpoint be $(x_2,y_2)$ and the mid - point be $M(x_m,y_m)$. Then $x_m=\frac{x_1 + x_2}{2}$ and $y_m=\frac{y_1 + y_2}{2}$. We can solve for $x_2$ and $y_2$: $x_2 = 2x_m-x_1$ and $y_2=2y_m - y_1$.

Step2: Solve for problem 7

Given $C(5,-6)$ as $(x_1,y_1)$ and $M(4,3)$ as $(x_m,y_m)$.
For the $x$ - coordinate of the other endpoint:
$x_2=2x_m - x_1=2\times4 - 5=8 - 5 = 3$.
For the $y$ - coordinate of the other endpoint:
$y_2=2y_m - y_1=2\times3-(-6)=6 + 6=12$.
The other endpoint is $(3,12)$.

Step3: Solve for problem 8

Given $D(-2,5)$ as $(x_1,y_1)$ and $M(-3,7)$ as $(x_m,y_m)$.
For the $x$ - coordinate of the other endpoint:
$x_2=2x_m - x_1=2\times(-3)-(-2)=-6 + 2=-4$.
For the $y$ - coordinate of the other endpoint:
$y_2=2y_m - y_1=2\times7 - 5=14 - 5 = 9$.
The other endpoint is $(-4,9)$.

Step4: Solve for problem 9

Given $D(8,0)$ as $(x_1,y_1)$ and $M(-2,9)$ as $(x_m,y_m)$.
For the $x$ - coordinate of the other endpoint:
$x_2=2x_m - x_1=2\times(-2)-8=-4 - 8=-12$.
For the $y$ - coordinate of the other endpoint:
$y_2=2y_m - y_1=2\times9-0 = 18$.
The other endpoint is $(-12,18)$.

Answer:

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