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'P' located at the point (x,y), and y = 2x. (ordinate = abcissa x 2) Distance of 'P' from (4,3) is square root of 10. (Distance)2 of 'P' from (4,3) is 10. ( x - 4 )2 + ( y - 3 )2 = 10 x2 - 8x + 16 + y2 - 6y + 9 = 10 x2 - 8x + y2 - 6y + 15 = 0 But y = 2x, so x2 - 8x +(2x)2 -6(2x) +15 = 0 5x2 - 20x + 15 = 0 ===> x2 - 4x + 3 = 0 ( x - 3 )( x - 1 ) = 0 x = 3 and x = 1 y = 6 and y = 2 The point 'P' can be either (3,6) or (1,2). Both points satisfy the given conditions.

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Q: A point P is at a distance of square root 10 units from the point 4 and 3 Find the coordinates of the point P if its ordinate is twice of its abscissa?
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