Question

show that the following point are the vertices of square (6,2) (2,1)(1,5)(5,6)​

Answer 1

Answer:

The given points are A (6,2), B(2,1), C(1,5) and D(5,6)

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

${ \red{AB = \sqrt{16 + 1}}}$

${ \huge{ \boxed{ \red{AB = \sqrt{17\: units} }}}}$

${ \pink{BC = \sqrt{ { (1 - 2)}^{2} + {(5 - 1)}^{2} } }}$

${ \pink{BC = \sqrt{ {( - 1)}^{2} + {( - 4)}^{2} } }}$

${ \pink{BC = \sqrt{1+ 16 } }}$

${ \huge{ \boxed{ \pink{BC = \sqrt{ 17\: units} }}}}$

${ \purple{CD = \sqrt{ {(5 - 1)}^{2} + {(6 - 5)}^{2} } }}$

${ \purple{CD = \sqrt{ {(4)}^{2} + {(1)}^{2} } }}$

${ \purple{CD = \sqrt{ 16 + 1 } }}$

${ \huge{ \boxed{ \purple{CD = \sqrt{ 17\: units } }}}}$

${ \blue{ DA= \sqrt{ {(5 - 6)}^{2} + {(6 - 2)}^{2} } }}$

${ \blue{ DA= \sqrt{ {(1)}^{2} + {(4)}^{2} } }}$

${ \blue{ DA= \sqrt{ 1 + 16 } }}$

${ \huge{ \boxed{ \blue{ DA= \sqrt{ 17 \: units } }}}}$

Therefore, AB =BC =CD =DA = 17 units

Also,

${ \green{AC = \sqrt{ {(1- 6)}^{2} + {(5 - 2)}^{2} } }}$

${ \green{AC = \sqrt{ {(- 5)}^{2} + {(3)}^{2} } }}$

${ \green{AC = \sqrt{ 25 + 9 } }}$

${ \huge{ \boxed { \green{AC = \sqrt{34\: units } }}}}$

${ \orange{ BD= \sqrt{ {(5- 2)}^{2} + {(6- 1)}^{2} } }}$

${ \orange{ BD= \sqrt{ {(3)}^{2} + {(5)}^{2} } }}$

${ \orange{ BD= \sqrt{ 9 + 25 } }}$

${ \huge{ \boxed { \orange{ BD= \sqrt{34\: units } }}}}$

Thus, diagonal AC = diagonal BD

Therefore, the given points from a square.

Step-by-step explanation:

Hope this helps you ✌️