Inquiry 1: When k = 1, the coordinate of the symmetrical point of point A about the straight line Y = X is (4,0), so the drawing is relatively simple;

When k = √ 3, the coordinate of the symmetrical point of point A about the straight line y = √ 3x is (2 √ 3,2).

Let point A be symmetrical about y = √ 3x line, and the coordinates of the symmetrical point are (x0, y0).

Then (y0+4)/2= √3(x0+0)/2 ①.

[(y0-4)/(x0-0)]*√3= - 1 ②

X0=2√3, y0=2 is used for ① and ② at the same time.

Question2: When k = 2, find the symmetrical point coordinates (8/5, -4/5) of point A about the straight line Y = 2x.

Let point A be symmetrical about the straight line Y = 2x, and the coordinates of the symmetrical point are (x0, y0).

Then (y0+4)/2= 2(x0+0)/2 ①.

[(y0-4)/(x0-0)]*2= - 1 ②

For both ① and ②, X0= 16/5 and y0= 12/5.

Application: Let point A be symmetrical about the straight line y = kx, and the coordinates of the symmetrical point are (x0, y0).

Then (y0+4)/2= k(x0+0)/2 ①.

[(y0-4)/(x0-0)]*k= - 1 ②

X0 = 8k/(k 2+ 1), y0 = 4 (k 2- 1)/(k 2+ 1) are used for both ① and ②.

So when k=m, the coordinates of point D are (8m/(m 2+ 1), 4 (m 2- 1)/(m 2+ 1)).

When k= 1/m, the coordinates of point G are (8m/(m 2+1), 4 (1-m 2)/(m 2+1).

Only when the abscissa of point D and point G are equal and the difference of ordinate is only 4, the quadrilateral AOGD is a diamond.

That is 4 (m2-1)/(m2+1)-4 (1-m2)/(m2+1) = 4.

The solution is m = √ 3, m = √1√ 3.

No graph is given, so judge whether m is positive or negative!