Question

directions: find the endpoint of a line segment when given an endpoint and the midpoint. 9. endpoint (-1,9) midpoint (-9,-10) 10. endpoint (2,5) midpoint (5,1) 11. endpoint (5,2) midpoint (-10,-2) 12.endpoint (9,-10) midpoint (4,8) 13. endpoint (-4,7) midpoint (10,3) 14.endpoint (-6,4) midpoint (4,8)

Explanation:

Step1: Recall mid - point formula

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

Step2: Solve for 9

Given $(x_1,y_1)=(-1,9)$ and $(m_x,m_y)=(-9,-10)$.
$x_2=2\times(-9)-(-1)=-18 + 1=-17$
$y_2=2\times(-10)-9=-20 - 9=-29$

Step3: Solve for 10

Given $(x_1,y_1)=(2,5)$ and $(m_x,m_y)=(5,1)$.
$x_2=2\times5 - 2=10 - 2 = 8$
$y_2=2\times1-5=2 - 5=-3$

Step4: Solve for 11

Given $(x_1,y_1)=(5,2)$ and $(m_x,m_y)=(-10,-2)$.
$x_2=2\times(-10)-5=-20 - 5=-25$
$y_2=2\times(-2)-2=-4 - 2=-6$

Step5: Solve for 12

Given $(x_1,y_1)=(9,-10)$ and $(m_x,m_y)=(4,8)$.
$x_2=2\times4-9=8 - 9=-1$
$y_2=2\times8-(-10)=16 + 10=26$

Step6: Solve for 13

Given $(x_1,y_1)=(-4,7)$ and $(m_x,m_y)=(10,3)$.
$x_2=2\times10-(-4)=20 + 4=24$
$y_2=2\times3-7=6 - 7=-1$

Step7: Solve for 14

Given $(x_1,y_1)=(-6,4)$ and $(m_x,m_y)=(4,8)$.
$x_2=2\times4-(-6)=8 + 6=14$
$y_2=2\times8-4=16 - 4=12$

Answer:

1. $(-17,-29)$
2. $(8,-3)$
3. $(-25,-6)$
4. $(-1,26)$
5. $(24,-1)$
6. $(14,12)$