Can somebody please help me with this question i attached the photo from khan academy.
Thank you!

Answer:
C = (12, 13)
Step-by-step explanation:
The centre of the rectangle is O.
Find the difference between the x and y-values of O and A:
[tex]O_x-A_x=\dfrac{21}{2}-9=\dfrac32[/tex]
[tex]O_y-A_y=\dfrac{19}{2}-6=\dfrac72[/tex]
Now add those difference to the coordinates of O to find C:
x value of C = [tex]O_x+\dfrac32=\dfrac{21}{2}+\dfrac32=\dfrac{24}{2}=12[/tex]
y value of C = [tex]O_y+\dfrac72=\dfrac{19}{2}+\dfrac72=\dfrac{26}{2}=13[/tex]
Therefore, the coordinates of C = (12, 13)
Answer:
Step-by-step explanation:
In rectangle the diagonals bisect each other. So,O is the midpoint of AC
[tex]Midpoint = \left(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2} \right)\\\\\\x_{1}=9 \ and \ y_{1}=6\\\\\left(\dfrac{9+x_{2}}{2},\dfrac{6+y_{2}}{2} \right) = \left (\dfrac{21}{2},\dfrac{19}{2} \right)[/tex]
Compare the x- coordinates and y- coordinates
[tex]\dfrac{9+x_{2}}{2}=\dfrac{21}{2} \ ; \ \dfrac{6+y_{2}}{2}=\dfrac{19}{2}\\\\\\9+x_{2}=\dfrac{21}{2}*2 \ ; \ 6 +y_{2}=\dfrac{19}{2}*2[/tex]
9 + x₂ = 21 ; 6 +y₂ = 19
x₂ = 21 -9 ; y₂ = 19 - 6
x₂ = 12 ; y₂ = 13
C(12 ,13)