Question

) Find the coordinates of the point A' which is the symmetric point

Answer 1

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 :

$C( \dfrac{x+3}{2}, \dfrac{y+2}{2})$

Also, product of slope will be -1 .( Since, they are parallel )

$\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1$

x = 2y - 1

So, $C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})$

Also, C satisfy given line :

$2\times ( \dfrac{2y + 2}{2}) + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}$

Also,

$x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}$

Therefore, the symmetric points is$A'(\dfrac{31}{5}, \dfrac{18}{5})$ .