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Each side of a square is of length 4. The centreof the square is (3, 7) and one ofits diagonalsis parallel to y x. Then co-ordinates of itsvertices are<C |
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Answer» Question-Each side of a square is of length 6 units and the centre of the square is (-1,2) .one of its diagnals is parallel to x+y=0. find the coordinates of the vertices of the square. solution:Square ABCD. Center = O (-1, 2).AB = BC = 6 units. So AC = 6√2 units. So AO = 3√2 units. Equation of diagonalAC: y + x = k , as it is parallel to x+y = 0Center O lies on AC: -1 + 2 = k => k =1 So Equation of AC: x + y = 1, or y = 1 - a. Let point A be (a, 1-a). Center O (-1, 2) is midpoint of AC.AO² = (a+1)² + (-a-1)² = (3√2)²=> a + 1 = + 3=> a = 2 or -4 As the diagonal AC has a slope -1, its inclination with x axis is 45°. So the sides AB and CD are parallel to x axis. AD and BC are parallel to y axis. A = (2, -1) , B(2-6, -1) or (-4,-1)D = (2, -1 + 6) or D= (2, 5), C(-4, 5) |
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