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**earboth** 1. The points B, D are the endpoints of the other diagonal. The diagonals intersect at the midpoint of AC: $\displaystyle M_{AC}\left(\frac{7+1}{2}~,~\frac{2+4}{2}\right)~\ implies~M_{AC}(4, 3)$

2. The points B, D are placed on the perpendicular bisector of AC:

Calculate the slope of AC:

$\displaystyle m_{AV} = \frac{4-2}{7-1} = -\frac13$

Therefore the slope of BD is:

$\displaystyle m_{BD} = -\frac1{m_{AC}}=3$

3. The points BD are connected by the straight line:

$\displaystyle y-3 = 3(x-4)~\implies~y = 3x-9$

4. The distances $\displaystyle |\overline{MA}|$ and $\displaystyle |\overline{MB}|$ must be equal. To get the coordinates of B you need a parallel to AC which have the distance $\displaystyle |\overline{MA}| = \sqrt{10}$ from AC.

Or:

B is the point of intersection between the perpendicular bisector of AC and the circle around M with radius $\displaystyle r = \sqrt{10}$:

$\displaystyle c: (x-4)^2 +(y-3)^2=10$

I've got: (5, 6), (3, 0)