UKPSC Mains PYQs

To find the coordinates of the remaining angular points of a square given the endpoints of its diagonal, we can use the properties of a square and the midpoint formula. Let’s denote the endpoints of the diagonal as ( A(3, 4) ) and ( B(1, 1) ).

First, we need to find the midpoint ( M ) of the diagonal ( AB ). The midpoint formula is given by: [ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) ] Substituting the coordinates of ( A ) and ( B ): [ M = \left( \frac{3 + 1}{2}, \frac{4 + 1}{2} \right) = \left( \frac{4}{2}, \frac{5}{2} \right) = \left( 2, 2.5 \right) ]

The midpoint ( M ) is also the center of the square. Since the diagonals of a square are equal and bisect each other at right angles, the other two vertices ( C ) and ( D ) of the square can be found by reflecting the endpoints ( A ) and ( B ) across the midpoint ( M ).

To find the coordinates of ( C ) and ( D ), we use the fact that the distance from ( M ) to ( A ) is the same as the distance from ( M ) to ( C ), and the distance from ( M ) to ( B ) is the same as the distance from ( M ) to ( D ).

The coordinates of ( C ) can be found by moving the same distance from ( M ) in the opposite direction of ( A ). The coordinates of ( D ) can be found by moving the same distance from ( M ) in the opposite direction of ( B ).

The vector from ( M ) to ( A ) is: [ \vec{MA} = (3 – 2, 4 – 2.5) = (1, 1.5) ] The vector from ( M ) to ( B ) is: [ \vec{MB} = (1 – 2, 1 – 2.5) = (-1, -1.5) ]

To find ( C ), we move from ( M ) in the direction opposite to ( A ): [ C = M – \vec{MA} = (2, 2.5) – (1, 1.5) = (1, 1) ]

To find ( D ), we move from ( M ) in the direction opposite to ( B ): [ D = M – \vec{MB} = (2, 2.5) – (-1, -1.5) = (3, 4) ]

However, we need to correct this as ( C ) and ( D ) should be distinct from ( A ) and ( B ). The correct coordinates of ( C ) and ( D ) are found by moving in the perpendicular direction to the diagonal. The perpendicular direction to ( \vec{MA} ) is ( (-1.5, 1) ) and to ( \vec{MB} ) is ( (1.5, -1) ).

Thus, the coordinates of ( C ) and ( D ) are: [ C = M + (1.5, -1) = (2 + 1.5, 2.5 – 1) = (3.5, 1.5) ] [ D = M – (1.5, -1) = (2 – 1.5, 2.5 + 1) = (0.5, 3.5) ]

Therefore, the coordinates of the remaining angular points of the square are ( (3.5, 1.5) ) and ( (0.5, 3.5) ).