Respuesta :
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 :
[tex]C( \dfrac{x+3}{2}, \dfrac{y+2}{2})[/tex]
Also, product of slope will be -1 .( Since, they are parallel )
[tex]\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1[/tex]
x = 2y - 1
So, [tex]C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})[/tex]
Also, C satisfy given line :
[tex]2\times ( \dfrac{2y + 2}{2}) + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}[/tex]
Also,
[tex]x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}[/tex]
Therefore, the symmetric points is[tex]A'(\dfrac{31}{5}, \dfrac{18}{5})[/tex] .
