Respuesta :
If point P is the midpoint of line АВ, then we can write the coordinates of the point P in this form:
[tex]\cfrac{X_A+X_B}{2}=X_P \ ; \ \ \ \ \ \cfrac{Y_A+Y_B}{2}=Y_P [/tex]
We know that [tex] X_A=-10 \ ; \ \ Y_A=8; \ \ X_P=-16 \ ; \ \ Y_P=6[/tex]
Now we can write two equations and solve them to find [tex]X_B\ \ and \ \ Y_B[/tex]
[tex]\frac{-10+X_B}{2}=-16 \\ -10+X_B=-16*2 \\ -10+X_B=-32 \\ X_B=-32+10 \\ \underline {X_B=-22}[/tex]
[tex]\frac{8+Y_B}{2}=6 \\ 8+Y_B=6*2 \\ 8+Y_B=12 \\ Y_B=12-8 \\ \underline {Y_B=4}[/tex]
Answer: B(-22; 4)
[tex]\cfrac{X_A+X_B}{2}=X_P \ ; \ \ \ \ \ \cfrac{Y_A+Y_B}{2}=Y_P [/tex]
We know that [tex] X_A=-10 \ ; \ \ Y_A=8; \ \ X_P=-16 \ ; \ \ Y_P=6[/tex]
Now we can write two equations and solve them to find [tex]X_B\ \ and \ \ Y_B[/tex]
[tex]\frac{-10+X_B}{2}=-16 \\ -10+X_B=-16*2 \\ -10+X_B=-32 \\ X_B=-32+10 \\ \underline {X_B=-22}[/tex]
[tex]\frac{8+Y_B}{2}=6 \\ 8+Y_B=6*2 \\ 8+Y_B=12 \\ Y_B=12-8 \\ \underline {Y_B=4}[/tex]
Answer: B(-22; 4)