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Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are $(9, -2)$ and $(1, 6)$ respectively.
(i) Find the co-ordinates of point P.
(ii) Find the length of the side of the square.
(i) Find the co-ordinates of point P.
(ii) Find the length of the side of the square.
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Sol. (i) Coordinates of P are $(\frac{9+1}{2}, \frac{-2+6}{2}) = (5,2)$ (1 Mark)
(ii) $2 AB^2 = BD^2$ (1/2 Mark)
$\Rightarrow 2 AB^2 = (9 – 1)^2 + (-2 – 6)^2$ (1/2 Mark)
$\Rightarrow AB = 8$ (1/2 Mark)
Hence, the length of the side of square is $8$ units.
(ii) $2 AB^2 = BD^2$ (1/2 Mark)
$\Rightarrow 2 AB^2 = (9 – 1)^2 + (-2 – 6)^2$ (1/2 Mark)
$\Rightarrow AB = 8$ (1/2 Mark)
Hence, the length of the side of square is $8$ units.