Question

Show that the points A (5, 6), B (1, 5),
C (2, 1) and D (6, 2) are the vertices of a
square ABCD.
PLZZ ANSWER..... ITS URGENT..... I WILL MARK BRAINLIEST! ​

Answer 1

Answer:

AB=

$\sqrt{ {(5 - 1)}^{2} + {(6 - 5)}^{2} } \\ = \sqrt{ {4}^{2} + {1}^{2} } = \sqrt{17}$

BC=

$\sqrt{ {(1 - 2)}^{2} + {(5 - 1)}^{2} } \\ = \sqrt{ { - 1}^{2} + {4}^{2} } \\ = \sqrt{17}$

CD=

$\sqrt{ {(2 - 6)}^{2} + {(1 - 2)}^{2} } \\ = \sqrt{ {( - 4)}^{2} + {( - 1)}^{2} } \\ = \sqrt{17}$

DA=

$\sqrt{ {(5 - 6)}^{2} + {(6 - 2)}^{2} } \\ = \sqrt{ {( - 1)}^{2} + {4}^{2} } \\ = \sqrt{17}$

As AB=BC=CD=DA are equal

all the sides are equal

so these are the vertices of square ABCD

Similarly check for diagonals we will get AC=BD