I don’t know the answer I can’t see the question that you are asking for
True because it wouldn’t have made since if it was false
Answer:
B
Step-by-step explanation:
It sounds most likely to the right answer
¯\_(ツ)_/¯
Let, coordinate of point A' is (x,y).
Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.
So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.
Now, their center is given by :
![C( \dfrac{x+3}{2}, \dfrac{y+2}{2})](https://tex.z-dn.net/?f=C%28%20%5Cdfrac%7Bx%2B3%7D%7B2%7D%2C%20%5Cdfrac%7By%2B2%7D%7B2%7D%29)
Also, product of slope will be -1 .( Since, they are parallel )
![\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1](https://tex.z-dn.net/?f=%5Cdfrac%7By-2%7D%7Bx-3%7D%20%5Ctimes%20-2%20%20%3D%20-1%5C%5C%5C%5C2y%20-%204%20%3D%20x%20-%203%5C%5C%5C%5C2y%20-%20x%20%3D%201)
x = 2y - 1
So, ![C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})](https://tex.z-dn.net/?f=C%28%20%5Cdfrac%7B2y%20-1%20%2B3%7D%7B2%7D%2C%20%5Cdfrac%7By%2B2%7D%7B2%7D%29%5C%5C%5C%5CC%28%20%5Cdfrac%7B2y%20%2B%202%7D%7B2%7D%2C%20%5Cdfrac%7By%2B2%7D%7B2%7D%29)
Also, C satisfy given line :
![2\times ( \dfrac{2y + 2}{2}) + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}](https://tex.z-dn.net/?f=2%5Ctimes%20%28%20%5Cdfrac%7B2y%20%2B%202%7D%7B2%7D%29%20%20%2B%20%5Cdfrac%7By%2B2%7D%7B2%7D%20%3D%2012%5C%5C%5C%5C4y%20%2B%204%20%2B%20y%20%2B%202%20%3D%2024%5C%5C%5C%5C5y%20%3D%2018%5C%5C%5C%5Cy%20%3D%20%5Cdfrac%7B18%7D%7B5%7D)
Also,
![x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}](https://tex.z-dn.net/?f=x%20%3D%202%5Ctimes%20%5Cdfrac%7B18%7D%7B5%20%7D%20-%201%5C%5C%5C%5Cx%20%3D%20%5Cdfrac%7B31%7D%7B5%7D)
Therefore, the symmetric points is
.